Appendix
Proof of Theorem 1
For any \(x, y_1,y_2, z_1, \ldots , z_{4}\), we have
$$\begin{aligned} & \langle A_{x} B_{1,y_1} B_{2,y_2} C_{1,z_1} C_{2,z_2} C_{3,z_3} C_{4,z_4}\rangle \\ &\quad= \sum \limits _{a, b_1,b_2, c_1,c_2, c_3, c_{4}} (-1)^{a+ b_1+ b_2+ c_1+c_2+c_3+ c_4} P(a, b_1,b_2, c_1,c_2, c_3, c_{4} | x, y_1,y_2, z_1, z_2, z_3, z_{4})\\ &\quad= \sum \limits _{a, b_1,b_2, c_1,c_2, c_3, c_{4}} (-1)^{a+ b_1+ b_2+ c_1+c_2+ c_3+ c_4} \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) P(a|x, \lambda _1,\lambda _2) P(b_1|y_1, \lambda _1, \lambda _3, \lambda _4)\\ &\quad \cdot P(b_2|y_2, \lambda _2, \lambda _5, \lambda _6) P(c_1|z_1, \lambda _{3}) P(c_2|z_2, \lambda _{4}) P(c_3|z_3, \lambda _{5}) P(c_4|z_4, \lambda _{6})\\ &\quad= \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) \left[\sum \limits _{a} (-1)^{a} P(a|x, \lambda _1, \lambda _2)\right] \left[\sum \limits _{b_1} (-1)^{b_1} P(b_1|y_1, \lambda _1, \lambda _3, \lambda _4)\right]\\ &\quad \cdot \left[\sum \limits _{b_2} (-1)^{b_2} P(b_2|y_2, \lambda _2, \lambda _5, \lambda _6)\right] \left[\sum \limits _{c_1} (-1)^{c_1} P(c_1|z_1, \lambda _{3})\right] \left[\sum \limits _{c_2} (-1)^{c_2} P(c_2|z_2, \lambda _{4})\right]\\ &\quad \cdot \left[\sum \limits _{c_3} (-1)^{c_3} P(c_3|z_3, \lambda _{5})\right] \left[\sum \limits _{c_4} (-1)^{c_4} P(c_4|z_4, \lambda _{6})\right]\\ &\quad= \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) \langle A_{x}\rangle _{\lambda _1, \lambda _2} \langle B_{1,y_1}\rangle _{\lambda _1, \lambda _3, \lambda _4} \langle B_{2,y_2}\rangle _{\lambda _{2}, \lambda _5, \lambda _6} \langle C_{1,z_{1}}\rangle _{\lambda _3} \langle C_{2,z_{2}}\rangle _{\lambda _4} \langle C_{3,z_{3}}\rangle _{\lambda _5} \langle C_{4,z_{4}}\rangle _{\lambda _6}, \end{aligned}$$
where
$$\begin{aligned} {\left\{ \begin{array}{ll}&\langle A_{x}\rangle _{\lambda _1, \lambda _2} = \sum \limits _{a} (-1)^{a} P(a|x, \lambda _1, \lambda _2),\\ &\langle B_{1,y_1}\rangle _{\lambda _1, \lambda _3, \lambda _4} = \sum \limits _{b_1} (-1)^{b_1} P(b_1|y_1, \lambda _1, \lambda _3, \lambda _4),\ \langle B_{2,y_2}\rangle _{\lambda _{2}, \lambda _5, \lambda _6} = \sum \limits _{b_2} (-1)^{b_2} P(b_2|y_2, \lambda _2, \lambda _5, \lambda _6),\\ &\langle C_{1,z_{1}}\rangle _{\lambda _3} = \sum \limits _{c_1} (-1)^{c_1} P(c_1|z_1, \lambda _{3}),\ \langle C_{2,z_{2}}\rangle _{\lambda _4} = \sum \limits _{c_2} (-1)^{c_2} P(c_2|z_2, \lambda _{4}),\\ &\langle C_{3,z_{3}}\rangle _{\lambda _5} = \sum \limits _{c_3} (-1)^{c_3} P(c_3|z_3, \lambda _{5}),\ \langle C_{4,z_{4}}\rangle _{\lambda _6} = \sum \limits _{c_4} (-1)^{c_4} P(c_4|z_4, \lambda _{6}).\end{array}\right. } \end{aligned}$$
Consequently, for any \(j\in \{1,\ldots , m\}\), we get
$$\begin{aligned} \begin{array}{rl} K_{m,j}=& \langle A_{j} B_{1,j} B_{2,j} (C_{1,j}+ C_{1,j+1}) (C_{2,j}+ C_{2,j+1}) (C_{3,j}+ C_{3,j+1}) (C_{4,j}+ C_{4,j+1})\rangle \\ =& \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) \langle A_{j}\rangle _{\lambda _1, \lambda _2} \langle B_{1,j}\rangle _{\lambda _1, \lambda _3, \lambda _4} \langle B_{2,j}\rangle _{\lambda _{2}, \lambda _5, \lambda _6} \left(\langle C_{1,j}\rangle _{\lambda _3} + \langle C_{1,j+1}\rangle _{\lambda _3} \right)\\ & \cdot \left(\langle C_{2,j}\rangle _{\lambda _4} + \langle C_{2,j+1}\rangle _{\lambda _4}\right) \left(\langle C_{3,j}\rangle _{\lambda _5} + \langle C_{3,j+1}\rangle _{\lambda _5}\right) \left(\langle C_{4,j}\rangle _{\lambda _6} + \langle C_{4,j+1}\rangle _{\lambda _6}\right). \end{array} \end{aligned}$$
This implies
$$\begin{aligned} \begin{array}{rl} &|K_{m,j}|\\ \le & \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) |\langle A_{j}\rangle _{\lambda _1, \lambda _2}| |\langle B_{1,j}\rangle _{\lambda _1, \lambda _3, \lambda _4}| |\langle B_{2,j}\rangle _{\lambda _{2}, \lambda _5, \lambda _6}| |\langle C_{1,j}\rangle _{\lambda _3} + \langle C_{1,j+1}\rangle _{\lambda _3}|\\ & \cdot |\langle C_{2,j}\rangle _{\lambda _4} + \langle C_{2,j+1}\rangle _{\lambda _4}| |\langle C_{3,j}\rangle _{\lambda _5} + \langle C_{3,j+1}\rangle _{\lambda _5}| |\langle C_{4,j}\rangle _{\lambda _6} + \langle C_{4,j+1}\rangle _{\lambda _6}|\\ \le & \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) |\langle C_{1,j}\rangle _{\lambda _3} + \langle C_{1,j+1}\rangle _{\lambda _3}| |\langle C_{2,j}\rangle _{\lambda _4} + \langle C_{2,j+1}\rangle _{\lambda _4}|\\ & \cdot |\langle C_{3,j}\rangle _{\lambda _5} + \langle C_{3,j+1}\rangle _{\lambda _5}| |\langle C_{4,j}\rangle _{\lambda _6} + \langle C_{4,j+1}\rangle _{\lambda _6}|\\ =& \left(\int \text{d}\lambda _3 P_3(\lambda _3) |\langle C_{1,j}\rangle _{\lambda _3}+ \langle C_{1,j+1}\rangle _{\lambda _3}|\right)\cdot \left(\int \text{d}\lambda _4 P_4(\lambda _4) |\langle C_{2,j}\rangle _{\lambda _4}+ \langle C_{2,j+1}\rangle _{\lambda _4}|\right)\\ & \cdot \left(\int \text{d}\lambda _5 P_5(\lambda _5) |\langle C_{3,j}\rangle _{\lambda _5}+ \langle C_{3,j+1}\rangle _{\lambda _5}|\right)\cdot \left(\int \text{d}\lambda _6 P_6(\lambda _6) |\langle C_{4,j}\rangle _{\lambda _6}+ \langle C_{4,j+1}\rangle _{\lambda _6}|\right), \end{array} \end{aligned}$$
where the second inequality follows from the facts that
$$\begin{aligned} |\langle A_{j}\rangle _{\lambda _1, \lambda _2}| \le 1,\ \ |\langle B_{1,j}\rangle _{\lambda _{1}, \lambda _3, \lambda _4}| \le 1,\ \ |\langle B_{2,j}\rangle _{\lambda _{2}, \lambda _5, \lambda _6}| \le 1. \end{aligned}$$
Since inequality \(\sum \limits _{k=1}^s (\prod \limits _{i=1}^t x_{i}^k)^{\frac{1}{t}} \le \prod \limits _{i=1}^t (\sum \limits _{k=1}^s x_{i}^k )^{\frac{1}{t}}\) holds for any \(x_i^k \ge 0\) (for example, see [16]), we can obtain that
$$\begin{aligned} \begin{array}{rl} & \sum \limits _{j=1}^{m} |K_{m,j}|^{\frac{1}{4}}\\ \le & \sum \limits _{j=1}^{m} \left[\left(\int \text{d}\lambda _3 P_3(\lambda _3) |\langle C_{1,j}\rangle _{\lambda _3}+ \langle C_{1,j+1}\rangle _{\lambda _3}|\right)\cdot \left(\int \text{d}\lambda _4 P_4(\lambda _4) |\langle C_{2,j}\rangle _{\lambda _4}+ \langle C_{2,j+1}\rangle _{\lambda _4}|\right)\right.\\ &\left. \cdot \left(\int \text{d}\lambda _5 P_5(\lambda _5) |\langle C_{3,j}\rangle _{\lambda _5}+ \langle C_{3,j+1}\rangle _{\lambda _5}|\right)\cdot \left(\int \text{d}\lambda _6 P_6(\lambda _6) |\langle C_{4,j}\rangle _{\lambda _6}+ \langle C_{4,j+1}\rangle _{\lambda _6}|\right)\right]^{\frac{1}{4}}\\ \le & \left(\int \text{d}\lambda _3 P_3(\lambda _3) \sum \limits _{j=1}^{m} |\langle C_{1,j}\rangle _{\lambda _3}+ \langle C_{1,j+1}\rangle _{\lambda _3}|\right)^{\frac{1}{4}} \cdot \left(\int \text{d}\lambda _4 P_4(\lambda _4) \sum \limits _{j=1}^{m} |\langle C_{2,j}\rangle _{\lambda _4}+ \langle C_{2,j+1}\rangle _{\lambda _4}|\right)^{\frac{1}{4}}\\ & \cdot \left(\int \text{d}\lambda _5 P_5(\lambda _5) \sum \limits _{j=1}^{m} |\langle C_{3,j}\rangle _{\lambda _5}+ \langle C_{3,j+1}\rangle _{\lambda _5}|\right)^{\frac{1}{4}} \cdot \left(\int \text{d}\lambda _6 P_6(\lambda _6) \sum \limits _{j=1}^{m} |\langle C_{4,j}\rangle _{\lambda _6}+ \langle C_{4,j+1}\rangle _{\lambda _6}|\right)^{\frac{1}{4}}. \end{array} \end{aligned}$$
Let us consider the expression \(\sum \limits _{j=1}^{m} |\langle C_{1,j}\rangle _{\lambda _3}+ \langle C_{1,j+1}\rangle _{\lambda _3}|\) firstly. Since
$$\begin{aligned} \begin{array}{rl} & |\langle C_{1,m}\rangle _{\lambda _3}- \langle C_{1,1}\rangle _{\lambda _3}| = |\langle C_{1,1}\rangle _{\lambda _3}- \langle C_{1,m}\rangle _{\lambda _3}|\\ \le & |\langle C_{1,1}\rangle _{\lambda _3}- \langle C_{1,2}\rangle _{\lambda _3}|+ |\langle C_{1,2}\rangle _{\lambda _3}- \langle C_{1,m}\rangle _{\lambda _3}|\\ \le & |\langle C_{1,1}\rangle _{\lambda _3}- \langle C_{1,2}\rangle _{\lambda _3}|+ |\langle C_{1,2}\rangle _{\lambda _3}- \langle C_{1,3}\rangle _{\lambda _3}|+ |\langle C_{1,3}\rangle _{\lambda _3}- \langle C_{1,m}\rangle _{\lambda _3}|\\ & \cdots \\ \le & |\langle C_{1,1}\rangle _{\lambda _3}- \langle C_{1,2}\rangle _{\lambda _3}|+ |\langle C_{1,2}\rangle _{\lambda _3}- \langle C_{1,3}\rangle _{\lambda _3}|+ \cdots + |\langle C_{1,m-1}\rangle _{\lambda _3}- \langle C_{1,m}\rangle _{\lambda _3}|, \end{array} \end{aligned}$$
it follows that
$$\begin{aligned} & \sum \limits _{j=1}^{m} |\langle C_{1,j}\rangle _{\lambda _3}+ \langle C_{1,j+1}\rangle _{\lambda _3}|\\ =& |\langle C_{1,1}\rangle _{\lambda _3}+ \langle C_{1,2}\rangle _{\lambda _3}|+ |\langle C_{1,2}\rangle _{\lambda _3}+ \langle C_{1,3}\rangle _{\lambda _3}|+ \cdots + |\langle C_{1,m-1}\rangle _{\lambda _3}+ \langle C_{1,m}\rangle _{\lambda _3}|+ |\langle C_{1,m}\rangle _{\lambda _3}- \langle C_{1,1}\rangle _{\lambda _3}|\\ \le & |\langle C_{1,1}\rangle _{\lambda _3}+ \langle C_{1,2}\rangle _{\lambda _3}|+
|\langle C_{1,2}\rangle _{\lambda _3}+ \langle C_{1,3}\rangle _{\lambda _3}|+ \cdots + |\langle C_{1,m-1}\rangle _{\lambda _3}+ \langle C_{1,m}\rangle _{\lambda _3}|\\ & + |\langle C_{1,1}\rangle _{\lambda _3}- \langle C_{1,2}\rangle _{\lambda _3}|+ |\langle C_{1,2}\rangle _{\lambda _3}- \langle C_{1,3}\rangle _{\lambda _3}|+ \cdots + |\langle C_{1,m-1}\rangle _{\lambda _3}- \langle C_{1,m}\rangle _{\lambda _3}|\\ =& (|\langle C_{1,1}\rangle _{\lambda _3}+ \langle C_{1,2}\rangle _{\lambda _3}|+ |\langle C_{1,1}\rangle _{\lambda _3}- \langle C_{1,2}\rangle _{\lambda _3}|)+ (|\langle C_{1,2}\rangle _{\lambda _3}+ \langle C_{1,3}\rangle _{\lambda _3}|+ |\langle C_{1,2}\rangle _{\lambda _3}- \langle C_{1,3}\rangle _{\lambda _3}|)+ \cdots \\ & +(|\langle C_{1,m-1}\rangle _{\lambda _3}+ \langle C_{1,m}\rangle _{\lambda _3}|+ |\langle C_{1,m-1}\rangle _{\lambda _3}- \langle C_{1,m}\rangle _{\lambda _3}|)\\ =& 2\max \{|\langle C_{1,1}\rangle _{\lambda _3}|, |\langle C_{1,2}\rangle _{\lambda _3}|\}+ 2\max \{|\langle C_{1,2}\rangle _{\lambda _3}|, |\langle C_{1,3}\rangle _{\lambda _3}|\}+ \cdots + 2\max \{|\langle C_{1,m-1}\rangle _{\lambda _3}|, |\langle C_{1,m}\rangle _{\lambda _3}|\}\\ \le & 2(m-1)= 2m-2, \end{aligned}$$
where the last inequality is due to \(|\langle C_{1,j}\rangle _{\lambda _3}| \le 1\) for all \(j\in \{1,\ldots ,m\}\).
Similarly, we can also get
$$\begin{aligned} \sum \limits _{j=1}^{m} |\langle C_{2,j}\rangle _{\lambda _4}+ \langle C_{2,j+1}\rangle _{\lambda _4}| \le 2m-2,\ \ \sum \limits _{j=1}^{m} |\langle C_{3,j}\rangle _{\lambda _5}+ \langle C_{3,j+1}\rangle _{\lambda _5}| \le 2m-2 \end{aligned}$$
and
$$\begin{aligned} \sum \limits _{j=1}^{m} |\langle C_{4,j}\rangle _{\lambda _6}+ \langle C_{4,j+1}\rangle _{\lambda _6}| \le 2m-2. \end{aligned}$$
Therefore, \(\sum \limits _{j=1}^{m} |K_{m,j}|^{\frac{1}{4}} \le 2\,m-2\). The proof is finished.□
Proof of Inequality (8). For any \(x,y_1, y_2,z_1,\ldots , z_4\), we have
$$\begin{aligned} \begin{array}{rl} & \langle A_{x} B_{1,y_1} B_{2,y_2} C_{1,z_1} C_{2,z_2} C_{3,z_3} C_{4,z_4}\rangle \\ =& \sum \limits _{a, b_1,b_2, c_1,\ldots , c_{4}} (-1)^{a+ b_1+ b_2+ c_1+\cdots + c_4} P(a, b_1,b_2, c_1,\ldots , c_{4} | x, y_1,y_2, z_1, \ldots , z_{4})\\ =& \sum \limits _{a, b_1,b_2, c_1,\ldots , c_{4}} (-1)^{a+ b_1+ b_2+ c_1+\cdots + c_4} \text{Tr}\left[\left(M_{a|x} \otimes M_{b_1|y_1} \otimes M_{b_2|y_2} \otimes M_{c_1|z_1} \otimes \cdots \otimes M_{c_{4}|z_{4}}\right)\rho \right]\\ =& \text{Tr}\left\{\left[\left(\sum \limits _{a} (-1)^{a} M_{a|x}\right) \otimes \left(\sum \limits _{b_1} (-1)^{b_1} M_{b_1|y_1}\right) \otimes \left(\sum \limits _{b_2} (-1)^{b_2} M_{b_2|y_2}\right) \otimes \left(\sum \limits _{c_1} (-1)^{c_1} M_{c_1|z_1}\right) \otimes \cdots \right.\right.\\ &\left.\left.\otimes \left(\sum \limits _{c_4} (-1)^{c_4} M_{c_4|z_4}\right)\right]\rho \right\}\\ =& \text{Tr}[(A_{x} \otimes B_{1,y_1} \otimes B_{2,y_2} \otimes C_{1,z_1} \otimes \cdots \otimes C_{4,z_4})\rho ]. \end{array} \end{aligned}$$
(15)
Consider the case of pure quantum state, i.e., \(\rho = |\psi \rangle \langle \psi |\) with
$$\begin{aligned} |\psi \rangle = |\psi \rangle _{AB_1} \otimes |\psi \rangle _{AB_2} \otimes |\psi \rangle _{B_1C_1} \otimes |\psi \rangle _{B_1C_2} \otimes |\psi \rangle _{B_2C_3} \otimes |\psi \rangle _{B_2C_4}. \end{aligned}$$
From the definition, we have
$$\begin{aligned} \begin{array}{ll} & \sum \limits _{j=1}^{m} |K_{m,j}|^{\frac{1}{4}}\\ &\quad = \sum \limits _{j=1}^{m} |\langle A_{j} \otimes B_{1,j}\otimes B_{2,j}\otimes (C_{1,j}+ C_{1,j+1})\otimes (C_{2,j}+ C_{2,j+1})\otimes (C_{3,j}+ C_{3,j+1})\otimes (C_{4,j}+ C_{4,j+1})\rangle |^{\frac{1}{4}}\\ &\quad = \sum \limits _{j=1}^{m} |\text{Tr}\{[A_{j}\otimes B_{1,j}\otimes B_{2,j}\otimes (C_{1,j}+ C_{1,j+1})\otimes (C_{2,j}+ C_{2,j+1})\otimes (C_{3,j}+ C_{3,j+1})\otimes (C_{4,j}+ C_{4,j+1})]|\psi \rangle \langle \psi |\} |^{\frac{1}{4}}\\ &\quad = \sum \limits _{j=1}^{m} |\langle \psi | [A_{j}\otimes B_{1,j}\otimes B_{2,j}\otimes (C_{1,j}+ C_{1,j+1})\otimes (C_{2,j}+ C_{2,j+1})\otimes (C_{3,j}+ C_{3,j+1})\otimes (C_{4,j}+ C_{4,j+1})]|\psi \rangle |^{\frac{1}{4}}\\&\quad = \sum \limits _{j=1}^{m} |\langle \psi | (A_{j}\otimes B_{1,j}\otimes B_{2,j}\otimes I^{C_1}\otimes I^{C_2}\otimes I^{C_3}\otimes I^{C_{4}})\\ &\quad \cdot [I^{A} \otimes I^{B_{1}} \otimes I^{B_{2}} \otimes (C_{1,j}+ C_{1,j+1})\otimes (C_{2,j}+ C_{2,j+1})\otimes (C_{3,j}+ C_{3,j+1})\otimes (C_{4,j}+ C_{4,j+1})]|\psi \rangle |^{\frac{1}{4}}\\ &\quad \le \sum \limits _{j=1}^{m} [\langle \psi | (A_{j}\otimes B_{1,j}\otimes B_{2,j}\otimes I^{C_1}\otimes I^{C_2}\otimes I^{C_3}\otimes I^{C_{4}})^2 |\psi \rangle ]^{\frac{1}{8}}\\ &\quad \cdot \{\langle \psi | [I^{A} \otimes I^{B_{1}} \otimes I^{B_{2}} \otimes (C_{1,j}+ C_{1,j+1})\otimes (C_{2,j}+ C_{2,j+1})\otimes (C_{3,j}+ C_{3,j+1})\otimes (C_{4,j}+ C_{4,j+1})]^2 |\psi \rangle \}^{\frac{1}{8}}\\&\quad = \sum \limits _{j=1}^{m} \{\langle \psi | [I^{A} \otimes I^{B_{1}} \otimes I^{B_{2}} \otimes (C_{1,j}+ C_{1,j+1})\otimes (C_{2,j}+ C_{2,j+1})\otimes (C_{3,j}+ C_{3,j+1})\otimes (C_{4,j}+ C_{4,j+1})]^2 |\psi \rangle \}^{\frac{1}{8}}\\ &\quad = \sum \limits _{j=1}^{m} \{\langle \psi | [I^{A} \otimes I^{B_{1}} \otimes I^{B_{2}} \otimes (C_{1,j}+ C_{1,j+1})^2\otimes (C_{2,j}+ C_{2,j+1})^2\otimes (C_{3,j}+ C_{3,j+1})^2\otimes (C_{4,j}+ C_{4,j+1})^2] |\psi \rangle \}^{\frac{1}{8}}, \end{array} \end{aligned}$$
where the inequality is due to Cauchy-Schwarz inequality \(|\langle v|w\rangle |^2 \le \langle v|v\rangle \langle w|w\rangle\) with the equality holding if and only if \(|v\rangle = \alpha |w\rangle\) for some scalar \(\alpha\), and the fifth equality follows from the facts \(A_{j}^2= B_{1,j}^2= B_{2,j}^2= I\) for \(j=1,\ldots ,m\). Note that, for any \(j\in \{1,\ldots , m\}\), it is true that
$$\begin{aligned} \begin{array}{rl} & \langle \psi | [I^{A} \otimes I^{B_{1}} \otimes I^{B_{2}} \otimes (C_{1,j}+ C_{1,j+1})^2\otimes (C_{2,j}+ C_{2,j+1})^2\otimes (C_{3,j}+ C_{3,j+1})^2\otimes (C_{4,j}+ C_{4,j+1})^2] |\psi \rangle \\ =& \langle \psi |_{B_1C_1} (C_{1,j}+C_{1,j+1})^2 |\psi \rangle _{B_1C_1} \langle \psi |_{B_1C_2} (C_{2,j}+ C_{2,j+1})^2 |\psi \rangle _{B_1C_2}\\ & \cdot \langle \psi |_{B_2C_3} (C_{3,j}+C_{3,j+1})^2 |\psi \rangle _{B_2C_3} \langle \psi |_{B_2C_4} (C_{4,j}+ C_{4,j+1})^2 |\psi \rangle _{B_2C_4}\\ =& \Vert (C_{1,j}+C_{1,j+1}) |\psi \rangle _{B_1C_1}\Vert ^2 \Vert (C_{2,j}+C_{2,j+1}) |\psi \rangle _{B_1C_2}\Vert ^2\Vert (C_{3,j}+C_{3,j+1}) |\psi \rangle _{B_2C_3}\Vert ^2 \Vert (C_{4,j}+C_{4,j+1}) |\psi \rangle _{B_2C_4}\Vert ^2\\ =& (\nu _{m,j}^1)^2 \cdot (\nu _{m,j}^2)^2 \cdot (\nu _{m,j}^3)^2 \cdot (\nu _{m,j}^4)^2, \end{array} \end{aligned}$$
where
$$\begin{aligned} \nu _{m,j}^1= & \Vert (C_{1,j}+ C_{1,j+1}) |\psi \rangle _{B_1C_1} \Vert ,\ \ \ \ \nu _{m,j}^2= \Vert (C_{2,j}+ C_{2,j+1}) |\psi \rangle _{B_1C_2} \Vert ,\\ \nu _{m,j}^3= & \Vert (C_{3,j}+ C_{3,j+1}) |\psi \rangle _{B_2C_3} \Vert ,\ \ \ \ \nu _{m,j}^4= \Vert (C_{4,j}+ C_{4,j+1}) |\psi \rangle _{B_2C_4} \Vert . \end{aligned}$$
Consequently,
$$\begin{aligned} \begin{array}{rl} \sum \limits _{j=1}^{m} |K_{m,j}|^{\frac{1}{4}} \le & \sum \limits _{j=1}^{m} \left[(\nu _{m,j}^1)^2 \cdot (\nu _{m,j}^2)^2 \cdot (\nu _{m,j}^3)^2 \cdot (\nu _{m,j}^4)^2\right]^{\frac{1}{8}}\\ =& \sum \limits _{j=1}^{m} \left[(\nu _{m,j}^1) \cdot (\nu _{m,j}^2) \cdot (\nu _{m,j}^3) \cdot (\nu _{m,j}^4)\right]^{\frac{1}{4}}\\ \le & \left[\sum \limits _{j=1}^{m} (\nu _{m,j}^1)\right]^{\frac{1}{4}} \cdot \left[\sum \limits _{j=1}^{m} (\nu _{m,j}^2)\right]^{\frac{1}{4}} \cdot \left[\sum \limits _{j=1}^{m} (\nu _{m,j}^3)\right]^{\frac{1}{4}} \cdot \left[\sum \limits _{j=1}^{m} (\nu _{m,j}^4)\right]^{\frac{1}{4}}\\ \le & 2m \cos (\frac{\pi }{2m}), \end{array} \end{aligned}$$
where the second inequality follows from \(\sum \limits _{k=1}^s (\prod \limits _{i=1}^t x_{i}^k)^{\frac{1}{t}} \le \prod \limits _{i=1}^t (\sum \limits _{k=1}^s x_{i}^k )^{\frac{1}{t}}\) \((\forall x_i^k \ge 0)\) with the equality holding if and only if \(x_{1}^k=\cdots = x_{t}^k\), and the last inequality is due to \(\sum \limits _{j=1}^{m} (\nu _{m,j}^{i})\le 2m \cos (\frac{\pi }{2m})\) \((i=1,2,3,4)\).□
Proof of Theorem 2
Assume \(\rho _{AB_1}= \sum \limits _u p_u^{AB_1} \rho _u^A\otimes \rho _u^{B_1}\), \(\rho _{AB_2}= \sum \limits _v p_v^{AB_2} \rho _v^A\otimes \rho _v^{B_2}\), \(\rho _{B_1C_1}= \sum \limits _w p_w^{B_1C_1} \rho _w^{B_1}\otimes \rho _w^{C_1}\), \(\rho _{B_1C_2}= \sum \limits _r p_r^{B_1C_2} \rho _r^{B_1}\otimes \rho _r^{C_2}\), \(\rho _{B_2C_3}= \sum \limits _s p_s^{B_2C_3} \rho _s^{B_2}\otimes \rho _s^{C_3}\), \(\rho _{B_2C_4}= \sum \limits _t p_t^{B_2C_4} \rho _t^{B_2}\otimes \rho _t^{C_4}\). Then
$$\begin{aligned} \begin{array}{rl} \rho =& \rho _{AB_1} \otimes \rho _{AB_2}\otimes \rho _{B_1C_1} \otimes \rho _{B_1C_2} \otimes \rho _{B_2C_3}\otimes \rho _{B_2C_4}\\ =& \left(\sum \limits _u p_u^{AB_1} \rho _u^A\otimes \rho _u^{B_1}\right)\otimes \left(\sum \limits _v p_v^{AB_2} \rho _v^A\otimes \rho _v^{B_2}\right)\\ & \otimes \left(\sum \limits _w p_w^{B_1C_1} \rho _w^{B_1}\otimes \rho _w^{C_1}\right) \otimes \left(\sum \limits _r p_r^{B_1C_2} \rho _r^{B_1}\otimes \rho _r^{C_2}\right)\\ &\otimes \left(\sum \limits _s p_s^{B_2C_3} \rho _s^{B_2}\otimes \rho _s^{C_3}\right)\otimes \left(\sum \limits _t p_t^{B_2C_4} \rho _t^{B_2}\otimes \rho _t^{C_4}\right)\\ =& \sum \limits _u p_u^{AB_1} \sum \limits _v p_v^{AB_2} \sum \limits _w p_w^{B_1C_1} \sum \limits _r p_r^{B_1C_2} \sum \limits _s p_s^{B_2C_3}\sum \limits _t p_t^{B_2C_4}\\ & (\rho _u^A\otimes \rho _u^{B_1}\otimes \rho _v^A\otimes \rho _v^{B_2} \otimes \rho _w^{B_1}\otimes \rho _w^{C_1}\otimes \rho _r^{B_1}\otimes \rho _r^{C_2}\\ & \otimes \rho _s^{B_2}\otimes \rho _s^{C_3}\otimes \rho _t^{B_2}\otimes \rho _t^{C_4}). \end{array} \end{aligned}$$
From Eq. (15), we have
$$\begin{aligned} \begin{array}{rl} &\langle A_{x} B_{1,y_1} B_{2,y_2} C_{1,z_1} C_{2,z_2} C_{3,z_3} C_{4,z_4}\rangle \\ =& {\rm Tr}[(A_{x} \otimes B_{1,y_1} \otimes B_{2,y_2} \otimes C_{1,z_1} \otimes C_{2,z_2} \otimes C_{3,z_3} \otimes C_{4,z_4})\rho ]\\ =& \sum \limits _u p_u^{AB_1} \sum \limits _v p_v^{AB_2} \sum \limits _w p_w^{B_1C_1} \sum \limits _r p_r^{B_1C_2}\sum \limits _s p_s^{B_2C_3}\sum \limits _t p_t^{B_2C_4}\\ & \{{\rm Tr}[(A_{x} \otimes B_{1,y_1} \otimes B_{2,y_2} \otimes C_{1,z_1} \otimes C_{2,z_2} \otimes C_{3,z_3} \otimes C_{4,z_4})(\rho _u^A\otimes \rho _u^{B_1}\otimes \rho _v^A\otimes \rho _v^{B_2}\\ & \otimes \rho _w^{B_1}\otimes \rho _w^{C_1}\otimes \rho _r^{B_1}\otimes \rho _r^{C_2}\otimes \rho _s^{B_2}\otimes \rho _s^{C_3}\otimes \rho _t^{B_2}\otimes \rho _t^{C_4})]\}\\ =& \sum \limits _u p_u^{AB_1} \sum \limits _v p_v^{AB_2} \sum \limits _w p_w^{B_1C_1} \sum \limits _r p_r^{B_1C_2}\sum \limits _s p_s^{B_2C_3}\sum \limits _t p_t^{B_2C_4}\\ & \{{\rm Tr}[A_{x}(\rho _u^A\otimes \rho _v^A)] \cdot {\rm Tr}[B_{1,y_1}(\rho _u^{B_1}\otimes \rho _w^{B_1}\otimes \rho _r^{B_1})] \cdot {\rm Tr}[B_{2,y_2}(\rho _v^{B_2}\otimes \rho _s^{B_2}\otimes \rho _t^{B_2})]\\ & \cdot {\rm Tr}[C_{1,z_1}(\rho _w^{C_1})] \cdot {\rm Tr}[C_{2,z_2}(\rho _r^{C_2})] \cdot {\rm Tr}[C_{3,z_3}(\rho _s^{C_3})] \cdot {\rm Tr}[C_{4,z_4}(\rho _t^{C_4})]\}. \end{array} \end{aligned}$$
Consequently, for any \(j\in \{1, \ldots ,m\}\),
$$\begin{aligned} \begin{array}{rl} K_{m,j}=& \langle A_{j} B_{1,j} B_{2,j} (C_{1,j}+ C_{1,j+1}) (C_{2,j}+ C_{2,j+1}) (C_{3,j}+ C_{3,j+1}) (C_{4,j}+ C_{4,j+1})\rangle \\ =& \sum \limits _u p_u^{AB_1} \sum \limits _v p_v^{AB_2} \sum \limits _w p_w^{B_1C_1} \sum \limits _r p_r^{B_1C_2}\sum \limits _s p_s^{B_2C_3}\sum \limits _t p_t^{B_2C_4} \{{\rm Tr}[A_{j}(\rho _u^A\otimes \rho _v^A)]\\ & \cdot {\rm Tr}[B_{1,j}(\rho _u^{B_1}\otimes \rho _w^{B_1}\otimes \rho _r^{B_1})] \cdot {\rm Tr}[B_{2,j}(\rho _v^{B_2}\otimes \rho _s^{B_2}\otimes \rho _t^{B_2})] \cdot {\rm Tr}[(C_{1,j}+ C_{1,j+1})(\rho _w^{C_1})]\\ & \cdot {\rm Tr}[(C_{2,j}+ C_{2,j+1})(\rho _r^{C_2})] \cdot {\rm Tr}[(C_{3,j}+ C_{3,j+1})(\rho _s^{C_3})] \cdot {\rm Tr}[(C_{4,j}+ C_{4,j+1})(\rho _t^{C_4})]\}, \end{array} \end{aligned}$$
which implies
$$\begin{aligned} \begin{array}{rl} |K_{m,j}|\le & \sum \limits _u p_u^{AB_1} \sum \limits _v p_v^{AB_2} \sum \limits _w p_w^{B_1C_1} \sum \limits _r p_r^{B_1C_2}\sum \limits _s p_s^{B_2C_3}\sum \limits _t p_t^{B_2C_4} \{|{\rm Tr}[A_{j}(\rho _u^A\otimes \rho _v^A)]|\\ & \cdot |{\rm Tr}[B_{1,j}(\rho _u^{B_1}\otimes \rho _w^{B_1}\otimes \rho _r^{B_1})]| \cdot |{\rm Tr}[B_{2,j}(\rho _v^{B_2}\otimes \rho _s^{B_2}\otimes \rho _t^{B_2})]| \cdot |{\rm Tr}[(C_{1,j}+ C_{1,j+1})(\rho _w^{C_1})]|\\ & \cdot |{\rm Tr}[(C_{2,j}+ C_{2,j+1})(\rho _r^{C_2})]| \cdot |{\rm Tr}[(C_{3,j}+ C_{3,j+1})(\rho _s^{C_3})]| \cdot |{\rm Tr}[(C_{4,j}+ C_{4,j+1})(\rho _t^{C_4})]|\}\\ \le & \sum \limits _u p_u^{AB_1} \sum \limits _v p_v^{AB_2} \sum \limits _w p_w^{B_1C_1} \sum \limits _r p_r^{B_1C_2}\sum \limits _s p_s^{B_2C_3}\sum \limits _t p_t^{B_2C_4}\\ & \{|{\rm Tr}[(C_{1,j}+ C_{1,j+1})(\rho _w^{C_1})]| \cdot |{\rm Tr}[(C_{2,j}+ C_{2,j+1})(\rho _r^{C_2})]|\\ & \cdot |{\rm Tr}[(C_{3,j}+ C_{3,j+1})(\rho _s^{C_3})]| \cdot |{\rm Tr}[(C_{4,j}+ C_{4,j+1})(\rho _t^{C_4})]|\}\\ =& \{\sum \limits _w p_w^{B_1C_1} |{\rm Tr}[(C_{1,j}+ C_{1,j+1})(\rho _w^{C_1})]|\} \cdot \{\sum \limits _r p_r^{B_1C_2} |{\rm Tr}[(C_{2,j}+ C_{2,j+1})(\rho _r^{C_2})]|\}\\ & \cdot \{\sum \limits _s p_s^{B_2C_3} |{\rm Tr}[(C_{3,j}+ C_{3,j+1})(\rho _s^{C_3})]|\} \cdot \{\sum \limits _t p_t^{B_2C_4} |{\rm Tr}[(C_{4,j}+ C_{4,j+1})(\rho _t^{C_4})]|\}, \end{array} \end{aligned}$$
where the second inequality is due to \(|{\rm Tr}[A_{j}(\rho _u^A\otimes \rho _v^A)]|\le \Vert A_{j}\Vert \cdot {\rm Tr}(\rho _u^A\otimes \rho _v^A) \le 1\), \(|{\rm Tr}[B_{1,j}(\rho _u^{B_1}\otimes \rho _w^{B_1}\otimes \rho _r^{B_1})]|\le 1\) and \(|{\rm Tr}[B_{2,j}(\rho _v^{B_2}\otimes \rho _s^{B_2}\otimes \rho _t^{B_2})]|\le 1\). Using inequality \(\sum \limits _{k=1}^s (\prod \limits _{i=1}^t x_{i}^k)^{\frac{1}{t}} \le \prod \limits _{i=1}^t (\sum \limits _{k=1}^s x_{i}^k )^{\frac{1}{t}}\) \((\forall x_i^k \ge 0)\), we get
$$\begin{aligned} \begin{array}{rl} \sum \limits _{j=1}^{m}|K_{m,j}|^{\frac{1}{4}} \le & \left\{\sum \limits _{j=1}^{m}\sum \limits _w p_w^{B_1C_1} |{\rm Tr}\left[(C_{1,j}+ C_{1,j+1})(\rho _w^{C_1})\right]|\right\}^{\frac{1}{4}} \cdot \left\{\sum \limits _{j=1}^{m}\sum \limits _r p_r^{B_1C_2} |{\rm Tr}\left[(C_{2,j}+ C_{2,j+1})(\rho _r^{C_2})\right]|\right\}^{\frac{1}{4}}\\ & \cdot \left\{\sum \limits _{j=1}^{m}\sum \limits _s p_s^{B_2C_3} |{\rm Tr}\left[(C_{3,j}+ C_{3,j+1})(\rho _s^{C_3})\right]|\right\}^{\frac{1}{4}} \cdot \left\{\sum \limits _{j=1}^{m}\sum \limits _t p_t^{B_2C_4} |{\rm Tr}\left[(C_{4,j}+ C_{4,j+1})(\rho _t^{C_4})\right]|\right\}^{\frac{1}{4}}. \end{array} \end{aligned}$$
We firstly consider the term \(\sum \limits _{j=1}^{m}\sum \limits _w p_w^{B_1C_1} |{\rm Tr}[(C_{1,j}+ C_{1,j+1})(\rho _w^{C_1})]|\). Since
$$\begin{aligned} \begin{array}{rl} & \sum \limits _{j=1}^{m} |{\rm Tr}[(C_{1,j}+ C_{1,j+1})(\rho _w^{C_1})]|\\ = & |{\rm Tr}[C_{1,1}(\rho _w^{C_1})]+ {\rm Tr}[C_{1,2}(\rho _w^{C_1})]|+ |{\rm Tr}[C_{1,2}(\rho _w^{C_1})]+ {\rm Tr}[C_{1,3}(\rho _w^{C_1})]|+\cdots \\ & +|{\rm Tr}[C_{1,m-1}(\rho _w^{C_1})]+ {\rm Tr}[C_{1,m}(\rho _w^{C_1})]|+ |{\rm Tr}[C_{1,m}(\rho _w^{C_1})]- {\rm Tr}[C_{1,1}(\rho _w^{C_1})]|\\ \le & |{\rm Tr}[C_{1,1}(\rho _w^{C_1})]+ {\rm Tr}[C_{1,2}(\rho _w^{C_1})]|+ |{\rm Tr}[C_{1,2}(\rho _w^{C_1})]+ {\rm Tr}[C_{1,3}(\rho _w^{C_1})]|+\cdots \\ & +|{\rm Tr}[C_{1,m-1}(\rho _w^{C_1})]+ {\rm Tr}[C_{1,m}(\rho _w^{C_1})]|+ \{|{\rm Tr}[C_{1,1}(\rho _w^{C_1})]- {\rm Tr}[C_{1,2}(\rho _w^{C_1})]|\\ & +|{\rm Tr}[C_{1,2}(\rho _w^{C_1})]- {\rm Tr}[C_{1,3}(\rho _w^{C_1})]|+\cdots +|{\rm Tr}[C_{1,m-1}(\rho _w^{C_1})]- {\rm Tr}[C_{1,m}(\rho _w^{C_1})]|\}\\ =& \{|{\rm Tr}[C_{1,1}(\rho _w^{C_1})]+ {\rm Tr}[C_{1,2}(\rho _w^{C_1})]|+ |{\rm Tr}[C_{1,1}(\rho _w^{C_1})]- {\rm Tr}[C_{1,2}(\rho _w^{C_1})]|\}\\ & +\{|{\rm Tr}[C_{1,2}(\rho _w^{C_1})]+ {\rm Tr}[C_{1,3}(\rho _w^{C_1})]|+ |{\rm Tr}[C_{1,2}(\rho _w^{C_1})]- {\rm Tr}[C_{1,3}(\rho _w^{C_1})]|\}+\cdots \\ & +\{|{\rm Tr}[C_{1,m-1}(\rho _w^{C_1})]+ {\rm Tr}[C_{1,m}(\rho _w^{C_1})]|+ |{\rm Tr}[C_{1,m-1}(\rho _w^{C_1})]- {\rm Tr}[C_{1,m}(\rho _w^{C_1})]|\}\\ =& 2\max \{|{\rm Tr}[C_{1,1}(\rho _w^{C_1})]|, |{\rm Tr}[C_{1,2}(\rho _w^{C_1})]|\}+ 2\max \{|{\rm Tr}[C_{1,2}(\rho _w^{C_1})]|, |{\rm Tr}[C_{1,3}(\rho _w^{C_1})]|\}\\ & +\cdots + 2\max \{|{\rm Tr}[C_{1,m-1}(\rho _w^{C_1})]|, |{\rm Tr}[C_{1,m}(\rho _w^{C_1})]|\}\\ \le & 2(m-1),
\end{array}\end{aligned}$$
where the last inequality follows from \(|{\rm Tr}[C_{1,j}(\rho _w^{C_1})]|\le \Vert C_{1,j}\Vert \cdot {\rm Tr}(\rho _w^{C_1}) \le 1\), \(\forall j\in \{1,\ldots , m\}\), then we have
$$\begin{aligned} \begin{array}{rl} &\sum \limits _{j=1}^{m}\sum \limits _w p_w^{B_1C_1} |{\rm Tr}[(C_{1,j}+ C_{1,j+1})(\rho _w^{C_1})]|\\ =& \sum \limits _w p_w^{B_1C_1}\sum \limits _{j=1}^{m} |{\rm Tr}[(C_{1,j}+ C_{1,j+1})(\rho _w^{C_1})]|\le 2(m-1). \end{array} \end{aligned}$$
Similarly, we obtain
$$\begin{aligned} & \sum \limits _{j=1}^{m}\sum \limits _r p_r^{B_1C_2} |{\rm Tr}[(C_{2,j}+ C_{2,j+1})(\rho _r^{C_2})]|\le 2(m-1),\\ & \sum \limits _{j=1}^{m}\sum \limits _s p_s^{B_2C_3} |{\rm Tr}[(C_{3,j}+ C_{3,j+1})(\rho _s^{C_3})]|\le 2(m-1),\\ & \sum \limits _{j=1}^{m}\sum \limits _t p_t^{B_2C_4} |{\rm Tr}[(C_{4,j}+ C_{4,j+1})(\rho _t^{C_4})]|\le 2(m-1). \end{aligned}$$
Therefore, \(\sum \limits _{j=1}^{m}|K_{m,j}|^{\frac{1}{4}}\le 2(m-1)\).□
Proof of Theorem 3
For any \(x, y_1,y_2, z_1, \ldots , z_{4}\), we have
$$\begin{aligned} \begin{array}{rl} & \langle A_{x} B_{1,y_1} B_{2,y_2} C_{1,z_1} C_{2,z_2} C_{3,z_3} C_{4,z_4}\rangle \\ =& \sum \limits _{a, b_1,b_2, c_1,c_2, c_3, c_{4}} (-1)^{a+ b_1+ b_2+ c_1+c_2+ c_3+ c_4} P(a, b_1,b_2, c_1,c_2, c_3, c_{4} | x, y_1,y_2, z_1, z_2, z_3, z_{4})\\ =& \sum \limits _{a, b_1,b_2, c_1,c_2, c_3, c_{4}} (-1)^{a+ b_1+ b_2+ c_1+c_2+ c_3+ c_4} \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) P(a|x, \lambda _1,\lambda _2) P(b_1|y_1, \lambda _1, \lambda _3, \lambda _4)\\ & \cdot P(b_2|y_2, \lambda _2, \lambda _5, \lambda _6) P(c_1|z_1, \lambda _{3}) P(c_2|z_2, \lambda _{4}) P(c_3|z_3, \lambda _{5}) P(c_4|z_4, \lambda _{6})\\ =& \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) [\sum \limits _{a} (-1)^{a} P(a|x, \lambda _1, \lambda _2)] [\sum \limits _{b_1} (-1)^{b_1} P(b_1|y_1, \lambda _1, \lambda _3, \lambda _4)]\\ & \cdot [\sum \limits _{b_2} (-1)^{b_2} P(b_2|y_2, \lambda _2, \lambda _5, \lambda _6)] [\sum \limits _{c_1} (-1)^{c_1} P(c_1|z_1, \lambda _{3})] [\sum \limits _{c_2} (-1)^{c_2} P(c_2|z_2, \lambda _{4})]\\ & \cdot [\sum \limits _{c_3} (-1)^{c_3} P(c_3|z_3, \lambda _{5})] [\sum \limits _{c_4} (-1)^{c_4} P(c_4|z_4, \lambda _{6})]\\ =& \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) \langle A_{x}\rangle _{\lambda _1, \lambda _2} \langle B_{1,y_1}\rangle _{\lambda _1, \lambda _3, \lambda _4} \langle B_{2,y_2}\rangle _{\lambda _{2}, \lambda _5, \lambda _6} \langle C_{1,z_{1}}\rangle _{\lambda _3} \langle C_{2,z_{2}}\rangle _{\lambda _4} \langle C_{3,z_{3}}\rangle _{\lambda _5} \langle C_{4,z_{4}}\rangle _{\lambda _6}, \end{array} \end{aligned}$$
where
$$\begin{aligned} {\left\{ \begin{array}{ll}&\langle A_{x}\rangle _{\lambda _1, \lambda _2} = \sum \limits _{a} (-1)^{a} P(a|x, \lambda _1, \lambda _2),\\ &\langle B_{1,y_1}\rangle _{\lambda _1, \lambda _3, \lambda _4} = \sum \limits _{b_1} (-1)^{b_1} P(b_1|y_1, \lambda _1, \lambda _3, \lambda _4),\ \langle B_{2,y_2}\rangle _{\lambda _{2}, \lambda _5, \lambda _6} = \sum \limits _{b_2} (-1)^{b_2} P(b_2|y_2, \lambda _2, \lambda _5, \lambda _6),\\ &\langle C_{1,z_{1}}\rangle _{\lambda _3} = \sum \limits _{c_1} (-1)^{c_1} P(c_1|z_1, \lambda _{3}),\ \langle C_{2,z_{2}}\rangle _{\lambda _4} = \sum \limits _{c_2} (-1)^{c_2} P(c_2|z_2, \lambda _{4}),\\ &\langle C_{3,z_{3}}\rangle _{\lambda _5} = \sum \limits _{c_3} (-1)^{c_3} P(c_3|z_3, \lambda _{5}),\ \langle C_{4,z_{4}}\rangle _{\lambda _6} = \sum \limits _{c_4} (-1)^{c_4} P(c_4|z_4, \lambda _{6}).\end{array}\right. } \end{aligned}$$
Then, for any \(j\in \{1,\ldots , 2^{m-1}\}\), one has
$$\begin{aligned} \begin{array}{rl} I_{m,j} =& \langle A_{j} B_{1,j} B_{2,j} [\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}] [\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}] [\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}] [\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}]\rangle \\ =& \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) \langle A_{j}\rangle _{\lambda _1, \lambda _2} \langle B_{1,j}\rangle _{\lambda _1, \lambda _3, \lambda _4} \langle B_{2,j}\rangle _{\lambda _{2}, \lambda _5, \lambda _6} [\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j} \langle C_{1,z_{1}}\rangle _{\lambda _3}]\\ & \cdot [\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j} \langle C_{2,z_{2}}\rangle _{\lambda _4}] [\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j} \langle C_{3,z_{3}}\rangle _{\lambda _5}] [\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j} \langle C_{4,z_{4}}\rangle _{\lambda _6}], \end{array} \end{aligned}$$
which implies
$$\begin{aligned} \begin{array}{rl} |I_{m,j}| \le & \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) |\langle A_{j}\rangle _{\lambda _1, \lambda _2}| |\langle B_{1,j}\rangle _{\lambda _1, \lambda _3, \lambda _4}| |\langle B_{2,j}\rangle _{\lambda _{2}, \lambda _5, \lambda _6}| |\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j} \langle C_{1,z_{1}}\rangle _{\lambda _3}|\\ & \cdot |\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j} \langle C_{2,z_{2}}\rangle _{\lambda _4}| |\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j} \langle C_{3,z_{3}}\rangle _{\lambda _5}| |\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j} \langle C_{4,z_{4}}\rangle _{\lambda _6}|\\ \le & \int \cdots \int \text{d}\lambda _1 \cdots \text{d}\lambda _{6} P_1(\lambda _1) \cdots P_6(\lambda _6) |\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j} \langle C_{1,z_{1}}\rangle _{\lambda _3}| |\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j} \langle C_{2,z_{2}}\rangle _{\lambda _4}|\\ & \cdot |\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j} \langle C_{3,z_{3}}\rangle _{\lambda _5}| |\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j} \langle C_{4,z_{4}}\rangle _{\lambda _6}|\\ =& [\int \text{d}\lambda _3 P_3(\lambda _3) |\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j} \langle C_{1,z_{1}}\rangle _{\lambda _3}|] \cdot [\int \text{d}\lambda _{4} P_4(\lambda _4) |\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j} \langle C_{2,z_{2}}\rangle _{\lambda _4}|]\\ & \cdot [\int \text{d}\lambda _{5} P_5(\lambda _5) |\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j} \langle C_{3,z_{3}}\rangle _{\lambda _5}|] \cdot [\int \text{d}\lambda _{6} P_6(\lambda _6) |\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j} \langle C_{4,z_{4}}\rangle _{\lambda _6}|], \end{array} \end{aligned}$$
where the second inequality follows from the facts that \(|\langle A_{j}\rangle _{\lambda _1, \lambda _2}| \le 1\), \(|\langle B_{1,j}\rangle _{\lambda _{1}, \lambda _3, \lambda _4}| \le 1\), \(|\langle B_{2,j}\rangle _{\lambda _{2}, \lambda _5, \lambda _6}| \le 1\). Using inequality \(\sum \limits _{k=1}^s (\prod \limits _{i=1}^t x_{i}^k)^{\frac{1}{t}} \le \prod \limits _{i=1}^t (\sum \limits _{k=1}^s x_{i}^k )^{\frac{1}{t}}\) \((\forall x_{i}^k\ge 0)\) again, we have
$$\begin{aligned} \begin{array}{rl} & \sum \limits _{j=1}^{2^{m-1}} |I_{m,j}|^{\frac{1}{4}}\\ \le & \sum \limits _{j=1}^{2^{m-1}} \{ [\int \text{d}\lambda _3 P_3(\lambda _3) |\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j} \langle C_{1,z_{1}}\rangle _{\lambda _3}|] \cdot [\int \text{d}\lambda _{4} P_4(\lambda _4) |\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j} \langle C_{2,z_{2}}\rangle _{\lambda _4}|]\\ & \cdot [\int \text{d}\lambda _{5} P_5(\lambda _5) |\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j} \langle C_{3,z_{3}}\rangle _{\lambda _5}|] \cdot [\int \text{d}\lambda _{6} P_6(\lambda _6) |\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j} \langle C_{4,z_{4}}\rangle _{\lambda _6}|] \}^{\frac{1}{4}}\\ \le & [\int \text{d}\lambda _3 P_3(\lambda _3) \sum \limits _{j=1}^{2^{m-1}} |\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j} \langle C_{1,z_{1}}\rangle _{\lambda _3}|] ^{\frac{1}{4}} \cdot [\int \text{d}\lambda _{4} P_4(\lambda _4) \sum \limits _{j=1}^{2^{m-1}} |\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j} \langle C_{2,z_{2}}\rangle _{\lambda _4}|] ^{\frac{1}{4}}\\ & \cdot [\int \text{d}\lambda _{5} P_5(\lambda _5) \sum \limits _{j=1}^{2^{m-1}} |\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j} \langle C_{3,z_{3}}\rangle _{\lambda _5}|] ^{\frac{1}{4}} \cdot [\int \text{d}\lambda _{6} P_6(\lambda _6) \sum \limits _{j=1}^{2^{m-1}} |\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j} \langle C_{4,z_{4}}\rangle _{\lambda _6}|] ^{\frac{1}{4}}\\ =& \sum \limits _{i=0}^{[ \frac{m}{2}]} (m-2i) \text{C}^i_m, \end{array} \end{aligned}$$
where the last equality is owing to
$$\begin{aligned} \begin{array}{rl}&\sum \limits _{j=1}^{2^{m-1}} |\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j} \langle C_{1,z_{1}}\rangle _{\lambda _3}| = \sum \limits _{j=1}^{2^{m-1}} |\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j} \langle C_{2,z_{2}}\rangle _{\lambda _4}|\\ =& \sum \limits _{j=1}^{2^{m-1}} |\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j} \langle C_{3,z_{3}}\rangle _{\lambda _5}| = \sum \limits _{j=1}^{2^{m-1}} |\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j} \langle C_{4,z_{4}}\rangle _{\lambda _6}| = \sum \limits _{i=0}^{[ \frac{m}{2}]}(m-2i) \text{C}^i_m.\end{array} \end{aligned}$$
Thus we complete the proof. \(\hfill\square\)
Proof of Inequality (12).
Consider the case of pure state, i.e.,
$$\begin{aligned} \rho = |\psi \rangle \langle \psi |\ \ \text{with}\ \ |\psi \rangle = |\psi \rangle _{AB_1} \otimes |\psi \rangle _{AB_2} \otimes |\psi \rangle _{B_1C_1} \otimes |\psi \rangle _{B_1C_2} \otimes |\psi \rangle _{B_2C_3} \otimes |\psi \rangle _{B_2C_4}. \end{aligned}$$
From the definition, we have
$$\begin{aligned} & \sum \limits _{j=1}^{2^{m-1}} |I_{m,j}|^{\frac{1}{4}}\\ =& \sum \limits _{j=1}^{2^{m-1}} |\langle A_{j}\otimes B_{1,j}\otimes B_{2,j}\otimes \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right]\otimes \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right]\otimes \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right]\otimes \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right]\rangle |^{\frac{1}{4}}\\ =& \sum \limits _{j=1}^{2^{m-1}} |\text{Tr}\left\{[A_{j}\otimes B_{1,j}\otimes B_{2,j}\otimes \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right]\otimes \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right]\otimes \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right]\right.\\ &\left. \otimes \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right]]|\psi \rangle \langle \psi |\right\} |^{\frac{1}{4}}\\ =& \sum \limits _{j=1}^{2^{m-1}} |\langle \psi | \left\{A_{j}\otimes B_{1,j}\otimes B_{2,j}\otimes \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right]\otimes \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right]\otimes \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right]\right.\\ &\left. \otimes \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right]\right\}|\psi \rangle |^{\frac{1}{4}}\\ =& \sum \limits _{j=1}^{2^{m-1}} |\langle \psi | (A_{j}\otimes B_{1,j}\otimes B_{2,j} \otimes I^{C_1} \otimes I^{C_2} \otimes I^{C_3} \otimes I^{C_4})\\ & \cdot \{ I^{A} \otimes I^{B_1}\otimes I^{B_2} \otimes \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right]\otimes \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right]\otimes \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right]\otimes \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right]\}|\psi \rangle |^{\frac{1}{4}}\\ \le & \sum \limits _{j=1}^{2^{m-1}} [\langle \psi | (A_{j}\otimes B_{1,j}\otimes B_{2,j} \otimes I^{C_1} \otimes I^{C_2} \otimes I^{C_3} \otimes I^{C_4})^2 |\psi \rangle ]^{\frac{1}{8}}\\ & \cdot (\langle \psi | \{I^{A} \otimes I^{B_1}\otimes I^{B_2} \otimes \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right]\otimes \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right]\otimes \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right]\\ & \otimes \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right]\}^2 |\psi \rangle )^{\frac{1}{8}}\\ =& \sum \limits _{j=1}^{2^{m-1}} (\langle \psi | \{I^{A} \otimes I^{B_1}\otimes I^{B_2} \otimes \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right]\otimes \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right]\otimes \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right]\\ & \otimes \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right]\}^2 |\psi \rangle )^{\frac{1}{8}}\\ =& \sum \limits _{j=1}^{2^{m-1}} (\langle \psi | \left\{I^{A} \otimes I^{B_1}\otimes I^{B_2} \otimes \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right]^2\otimes \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right]^2\otimes \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right]^2\right. \\ &\left. \otimes \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right]^2\right\} |\psi \rangle )^{\frac{1}{8}}, \end{aligned}$$
where the above inequality is due to Cauchy-Schwarz inequality and the fifth equality follows from the fact \(A_{j}^2= B_{1,j}^2= B_{2,j}^2= I\) for \(j=1,\ldots ,2^{m-1}\). Note that, for any \(j\in \{1,\ldots , 2^{m-1}\}\), one has
$$\begin{aligned} & \left\langle \psi \left| \left\{I^{A} \otimes I^{B_{1}} \otimes I^{B_{2}} \otimes \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right]^2\otimes \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right]^2\otimes \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right]^2 \otimes \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right]^2\right\} \right|\psi \right\rangle \\ &\quad = \langle \psi |_{B_1C_1} \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right]^2 |\psi \rangle _{B_1C_1} \cdot \langle \psi |_{B_1C_2} \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right]^2 |\psi \rangle _{B_1C_2}\\ &\qquad \cdot \langle \psi |_{B_2C_3} \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right]^2 |\psi \rangle _{B_2C_3} \cdot \langle \psi |_{B_2C_4} \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right]^2 |\psi \rangle _{B_2C_4}\\ &\quad = \Vert \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right] |\psi \rangle _{B_1C_1}\Vert ^2 \cdot \Vert \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right] |\psi \rangle _{B_1C_2}\Vert ^2\\ &\qquad \cdot \Vert \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right] |\psi \rangle _{B_2C_3}\Vert ^2 \cdot \Vert \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right] |\psi \rangle _{B_2C_4}\Vert ^2\\ &\quad = (\omega _{m,j}^1)^2 \cdot (\omega _{m,j}^2)^2 \cdot (\omega _{m,j}^3)^2 \cdot (\omega _{m,j}^4)^2, \end{aligned}$$
where
$$\begin{aligned} \omega _{m,j}^1= & \Vert \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right] |\psi \rangle _{B_1C_1} \Vert ,\ \ \ \ \omega _{m,j}^2= \Vert \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right] |\psi \rangle _{B_1C_2} \Vert ,\\ \omega _{m,j}^3= & \Vert \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right] |\psi \rangle _{B_2C_3} \Vert ,\ \ \ \ \omega _{m,j}^4= \Vert \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right] |\psi \rangle _{B_2C_4} \Vert . \end{aligned}$$
So
$$\begin{aligned} \begin{array}{rl} \sum \limits _{j=1}^{2^{m-1}} |I_{m,j}|^{\frac{1}{4}} \le & \sum \limits _{j=1}^{2^{m-1}} \left[(\omega _{m,j}^1)^2 \cdot (\omega _{m,j}^2)^2 \cdot (\omega _{m,j}^3)^2 \cdot (\omega _{m,j}^4)^2\right]^{\frac{1}{8}}\\ =& \sum \limits _{j=1}^{2^{m-1}} \left[(\omega _{m,j}^1) \cdot (\omega _{m,j}^2) \cdot (\omega _{m,j}^3) \cdot (\omega _{m,j}^4)\right]^{\frac{1}{4}}\\ \le & \left[\sum \limits _{j=1}^{2^{m-1}} (\omega _{m,j}^1)\right]^{\frac{1}{4}} \cdot \left[\sum \limits _{j=1}^{2^{m-1}} (\omega _{m,j}^2)\right]^{\frac{1}{4}} \cdot \left[\sum \limits _{j=1}^{2^{m-1}} (\omega _{m,j}^3)\right]^{\frac{1}{4}} \cdot \left[\sum \limits _{j=1}^{2^{m-1}} (\omega _{m,j}^4)\right]^{\frac{1}{4}}\\ \le & 2^{m-1} \sqrt{m}, \end{array} \end{aligned}$$
where the last inequality is due to \(\sum \limits _{j=1}^{2^{m-1}} (\omega _{m,j}^i)\le 2^{m-1}\sqrt{m}\) \((i=1,2,3,4)\). Hence \((\sum \limits _{j=1}^{2^{m-1}} |I_{m,j}|^{\frac{1}{4}})_Q\le 2^{m-1} \sqrt{m}\), as desired. \(\hfill\square\)
Proof of Theorem 4
Assume \(\rho _{AB_1}\), \(\rho _{AB_2}\), \(\rho _{B_1C_1}\), \(\rho _{B_1C_2}\), \(\rho _{B_2C_3}\), \(\rho _{B_2C_4}\) have the forms as in the proof of Theorem 2. Then, for any \(j\in \{1, \ldots ,2^{m-1}\}\), we have
$$\begin{aligned} \begin{array}{rl} I_{m,j}=& \langle A_{j}B_{1,j} B_{2,j} \left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right] \left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right] \left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right] \left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right]\rangle \\ =& \sum \limits _u p_u^{AB_1} \sum \limits _v p_v^{AB_2} \sum \limits _w p_w^{B_1C_1}\sum \limits _r p_r^{B_1C_2}\sum \limits _s p_s^{B_2C_3}\sum \limits _t p_t^{B_2C_4} \Bigg\{{\rm Tr}[A_{j}(\rho _u^A\otimes \rho _v^A)]\\& \cdot {\rm Tr}[B_{1,j}(\rho _u^{B_1}\otimes \rho _w^{B_1}\otimes \rho _r^{B_1})] \cdot {\rm Tr}[B_{2,j}(\rho _v^{B_2}\otimes \rho _s^{B_2}\otimes \rho _t^{B_2})] \cdot {\rm Tr}\left\{\left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right](\rho _w^{C_1})\right\}\\ & \cdot {\rm Tr}\left\{\left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right](\rho _r^{C_2})\right\} \cdot {\rm Tr}\left\{\left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right](\rho _s^{C_3})\right\} \cdot {\rm Tr}\left\{\left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right](\rho _t^{C_4})\right\}\Bigg\}, \end{array}\end{aligned}$$
which implies
$$\begin{aligned} \begin{array}{rl} |I_{m,j}|\le & \sum \limits _u p_u^{AB_1} \sum \limits _v p_v^{AB_2} \sum \limits _w p_w^{B_1C_1} \sum \limits _r p_r^{B_1C_2}\sum \limits _s p_s^{B_2C_3}\sum \limits _t p_t^{B_2C_4} \Bigg\{|{\rm Tr}[A_{j}(\rho _u^A\otimes \rho _v^A)]|\\ & \cdot |{\rm Tr}[B_{1,j}(\rho _u^{B_1}\otimes \rho _w^{B_1}\otimes \rho _r^{B_1})]| \cdot |{\rm Tr}[B_{2,j}(\rho _v^{B_2}\otimes \rho _s^{B_2}\otimes \rho _t^{B_2})]| \cdot |{\rm Tr}\left\{\left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right](\rho _w^{C_1})\right\}|\\ &\cdot |{\rm Tr}\left\{\left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right](\rho _r^{C_2})\right\}| \cdot |{\rm Tr}\left\{\left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right](\rho _s^{C_3})\right\}| \cdot |{\rm Tr}\left\{\left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right](\rho _t^{C_4})\right\}|\Bigg\}\\ \le & \sum \limits _u p_u^{AB_1} \sum \limits _v p_v^{AB_2} \sum \limits _w p_w^{B_1C_1} \sum \limits _r p_r^{B_1C_2}\sum \limits _s p_s^{B_2C_3}\sum \limits _t p_t^{B_2C_4}\\ &\Bigg\{|{\rm Tr}\left\{\left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right](\rho _w^{C_1})\right\}| \cdot |{\rm Tr}\left\{\left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right](\rho _r^{C_2})\right\}|\\ & \cdot |{\rm Tr}\left\{\left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right](\rho _s^{C_3})\right\}| \cdot |{\rm Tr}\left\{\left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right](\rho _t^{C_4})\right\}|\Bigg\}\\ =& \left\{\sum \limits _w p_w^{B_1C_1} |{\rm Tr}\left\{\left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right](\rho _w^{C_1})\right\}|\right\} \cdot \left\{\sum \limits _r p_r^{B_1C_2} |{\rm Tr}\left\{\left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right](\rho _r^{C_2})\right\}|\right\}\\ & \cdot \left\{\sum \limits _s p_s^{B_2C_3}|{\rm Tr}\left\{\left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right](\rho _s^{C_3})\right\}|\right\} \cdot \left\{\sum \limits _t p_t^{B_2C_4} |{\rm Tr}\left\{\left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right](\rho _t^{C_4})\right\}|\right\}.\end{array} \end{aligned}$$
By inequality \(\sum \limits _{k=1}^s (\prod \limits _{i=1}^t x_{i}^k)^{\frac{1}{t}} \le \prod \limits _{i=1}^t (\sum \limits _{k=1}^s x_{i}^k )^{\frac{1}{t}}\) \((\forall x_i^k \ge 0)\), we can obtain that
$$\begin{aligned} \begin{array}{rl} &\sum \limits _{j=1}^{2^{m-1}}|I_{m,j}|^{\frac{1}{4}}\\ \le & \left\{\sum \limits _{j=1}^{2^{m-1}}\sum \limits _w p_w^{B_1C_1} |{\rm Tr}\left\{\left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right](\rho _w^{C_1})\right\}|\right\}^{\frac{1}{4}} \cdot \left\{\sum \limits _{j=1}^{2^{m-1}}\sum \limits _r p_r^{B_1C_2} |{\rm Tr}\left\{\left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right](\rho _r^{C_2})\right\}|\right\}^{\frac{1}{4}}\\ & \cdot \left\{\sum \limits _{j=1}^{2^{m-1}}\sum \limits _s p_s^{B_2C_3} |{\rm Tr}\left\{\left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right](\rho _s^{C_3})\right\}|\right\}^{\frac{1}{4}} \cdot \left\{\sum \limits _{j=1}^{2^{m-1}}\sum \limits _t p_t^{B_2C_4} |{\rm Tr}\left\{\left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right](\rho _t^{C_4})\right\}|\right\}^{\frac{1}{4}}. \end{array} \end{aligned}$$
For the first term in the above expression, we have
$$\begin{aligned} \begin{array}{rl} &\sum \limits _{j=1}^{2^{m-1}}\sum \limits _w p_w^{B_1C_1} |{\rm Tr}\left\{\left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right](\rho _w^{C_1})\right\}|\\ =& \sum \limits _w p_w^{B_1C_1}\sum \limits _{j=1}^{2^{m-1}} |{\rm Tr}\left\{\left[\sum \limits _{z_{1}=1}^m (-1)^{\theta _{z_{1}}^j}C_{1,z_{1}}\right](\rho _w^{C_1})\right\}|\\ =& \sum \limits _w p_w^{B_1C_1} \left[\sum \limits _{i=0}^{\left[ \frac{m}{2}\right]}(m-2i) \text{C}^i_m\right]= \sum \limits _{i=0}^{\left[ \frac{m}{2}\right]}(m-2i) \text{C}^i_m. \end{array} \end{aligned}$$
Similarly, we have
$$\begin{aligned} & \sum \limits _{j=1}^{2^{m-1}}\sum \limits _r p_r^{B_1C_2} |{\rm Tr}\left\{\left[\sum \limits _{z_{2}=1}^m (-1)^{\theta _{z_{2}}^j}C_{2,z_{2}}\right](\rho _r^{C_2})\right\}|= \sum \limits _{i=0}^{\left[ \frac{m}{2}\right]}(m-2i) \text{C}^i_m,\\ & \sum \limits _{j=1}^{2^{m-1}}\sum \limits _s p_s^{B_2C_3} |{\rm Tr}\left\{\left[\sum \limits _{z_{3}=1}^m (-1)^{\theta _{z_{3}}^j}C_{3,z_{3}}\right](\rho _s^{C_3})\right\}|= \sum \limits _{i=0}^{\left[ \frac{m}{2}\right]}(m-2i) \text{C}^i_m,\\ & \sum \limits _{j=1}^{2^{m-1}}\sum \limits _t p_t^{B_2C_4} |{\rm Tr}\left\{\left[\sum \limits _{z_{4}=1}^m (-1)^{\theta _{z_{4}}^j}C_{4,z_{4}}\right](\rho _t^{C_4})\right\}|= \sum \limits _{i=0}^{\left[ \frac{m}{2}\right]}(m-2i) \text{C}^i_m. \end{aligned}$$
Therefore, \(\sum \limits _{j=1}^{2^{m-1}}|I_{m,j}|^{\frac{1}{4}}\le \sum \limits _{i=0}^{[ \frac{m}{2}]}(m-2i) \text{C}^i_m\). \(\hfill\square\)
