Pricing-decision analysis of green supply chain with two competitive manufacturers considering horizontal and vertical fairness concerns

Abstract

A pricing-decision analysis is a critical aspect of supply chain management since it directly affects manufacturers’ and retailers’ profits. The members of a supply chain all want to be treated properly during the pricing process, which means that they pay attention to the fairness of the profit distribution. Additionally, consumers are increasingly gravitating to green products as their awareness of green consumption grows. Thus, incorporating consumers’ green preferences into a supply chain, this paper investigates pricing decisions with two competitive manufacturers under horizontal and vertical fairness concerns and seeks the optimal degrees of product greenness, prices, profits, and utilities. The game-theoretical models with and without bidirectional fairness concerns are constructed and analyzed to identify the implications on pricing, profits, and utilities of competing manufacturers’ bidirectional fairness concerns and consumers’ green preferences. Then, we determine the decisional differences between the two designs using comparative analysis and numerical simulation. Finally, propositions, corollaries, and policy implications are derived. The results indicate that consumers’ green preferences and competition between manufacturers contribute to increasing the optimal pricing and retailer’s profit while harming manufacturers’ utilities and the supply chain’s profits under some conditions. The findings also demonstrate that horizontal and vertical fairness concerns generate different impacts on the product’s greenness degree and pricing. Still, they are detrimental to manufacturers’ utilities and supply chain profit while possessing negligible effects on retailer’s profit.

Appendix

Proof of Proposition 1

The manufacturers simultaneously and uncooperatively determine the wholesale price and product greening degree, and then the retailer determine the retail price based on the given wholesale price and greening degree. The Stackelberg game between the leader and the follower can be analyzed by backward induction procedure.

In this set-up, we solve for the retailer’s profit function first. We use 1 and 2 to represent the two manufacturers as well as their products, respectively. The first partial derivative of π_ri to p₁ and p₂ can be shown as ∂π_r/∂p₁ = a − 2p₁ + 2rp₂ − αrτ₂ + ατ₁ − rw₂ + w₁, ∂π_r/∂p₂ = a − 2p₂ + 2rp₁ − αrτ₁ + ατ₂ − rw₁ + w₂. The second partial derivatives of π_ri to p₁ and p₂ are ${\partial}^2{\pi}_r/\partial {p}_1^2=-2$, ${\partial}^2{\pi}_r/\partial {p}_2^2=-2$, and ∂²π_r/∂p₁∂p₂ = 2r; based on the assumption 0 < r < 1, we can derive 4 − 4r² > 0. Thus, the Hessian matrix is negative. Hence the retailer’s profit is jointly concave in p₁ and p₂. Equating the first partial derivative to 0, we get ${p}_i^{\ast }=\left(a+\left(1-r\right)\left(\alpha {\tau}_i+{w}_i\right)\right)/\left(2\left(1-r\right)\right)$, i = 1, 2.

Then, we substitute the  p_i^* into Eq. (2), and we have the first partial conditions which are $\partial {\pi}_{m1}/\partial {w}_1=\frac{1}{2}\left(a+c+\alpha {\tau}_1-\alpha r{\tau}_2-2{w}_1+r{w}_2\right)$, $\partial {\pi}_{m2}/\partial {w}_2=\frac{1}{2}\left(a+c+\alpha {\tau}_2-\alpha r{\tau}_1-2{w}_2+r{w}_1\right)$, $\partial {\pi}_{m1}/\partial {\tau}_1=\frac{1}{2}\left(\alpha {w}_1-\alpha c-2\beta {\tau}_1\right)$, and $\partial {\pi}_{m2}/\partial {\tau}_1=\frac{1}{2}\left(\alpha {w}_2-\alpha c-2\beta {\tau}_2\right)$. The second partial derivative ${\partial}^2{\pi}_{mi}/\partial {w}_i^2=-1$, ${\partial}^2{\pi}_{mi}/\partial {\tau}_i^2=-\beta$, and ∂²π_mi/∂w_i∂τ_i = α/2; hence, the Hessian matrix can be shown as $H=\left(\begin{array}{c}-1\\ {}\frac{\alpha }{2}\end{array}\right.\left.\begin{array}{c}\frac{\alpha }{2}\\ {}-\beta\end{array}\right)$, which is non-negative under condition 4β > α². Then the π_mi is jointly concave in w_i and τ_i. Equating the first partial derivative to 0, we can get ${\tau}_i^{NN^{\ast }}=\frac{\alpha \left[a-c\left(1-r\right)\right]}{2\beta \left(2-r\right)-{\alpha}^2\left(1-r\right)}$ and ${w}_i^{NN^{\ast }}=\frac{c\left[2\beta -{\alpha}^2\left(1-r\right)\right]+2 a\beta}{2\beta \left(2-r\right)-{\alpha}^2\left(1-r\right)}$. Replacing w_i and τ_i with ${w}_i^{NN^{\ast }}$ and ${\tau}_i^{NN^{\ast }}$ into p_i^* and the ${p}_i^{NN^{\ast }}$ can be obtained. Substituting ${w}_i^{NN^{\ast }}$, ${\tau}_i^{NN^{\ast }}$, and ${p}_i^{NN^{\ast }}$ into Eqs. (1)–(3), we can derive ${\pi}_r^{NN^{\ast }}$, ${\pi}_{mi}^{NN^{\ast }}$, and${\pi}^{NN^{\ast }}$.

Proof of Corollary 1

Based on the equilibrium solutions obtained in Proposition 1, we have $\frac{\partial {\tau}_i^{NN^{\ast }}}{\partial \alpha }=\frac{\alpha \left[a-c\left(1-r\right)\right]\left[2\beta \left(2-r\right)+{\alpha}^2\left(1-r\right)\right]}{{\left[2\beta \left(2-r\right)-{\alpha}^2\left(1-r\right)\right]}^2}>0$, $\frac{\partial {w}_i^{NN^{\ast }}}{\partial \alpha }=\frac{4\beta \alpha \left[a-c\left(1-r\right)\right]\left(1-r\right)}{{\left[2\beta \left(2-r\right)-{\alpha}^2\left(1-r\right)\right]}^2}>0$, and $\frac{\partial {p}_i^{NN^{\ast }}}{\partial \alpha }=\frac{2\beta \alpha \left[a-c\left(1-r\right)\right]\left(3-2r\right)}{{\left[2\beta \left(2-r\right)-{\alpha}^2\left(1-r\right)\right]}^2}>0$.

Proof of Corollary 2

Similar to above proof, the following properties hold as $\frac{\partial {\pi}_r^{NN^{\ast }}}{\partial \alpha }=\frac{8{\beta}^2\alpha {\left[a-c\left(1-r\right)\right]}^2}{{\left[2\beta \left(2-r\right)-{\alpha}^2\left(1-r\right)\right]}^3}>0$, $\frac{\partial {\pi}^{NN^{\ast }}}{\partial \alpha }=\frac{2\alpha \beta {\left[a-c\left(1-r\right)\right]}^2\left[2\beta \left(4-3r\right)-{\alpha}^2\left(1-r\right)\right]}{{\left[2\beta \left(2-r\right)-{\alpha}^2\left(1-r\right)\right]}^3}>0$, and $\frac{\partial {\pi}_{mi}^{NN^{\ast }}}{\partial \alpha }=\frac{\beta \alpha {\left[a-c\left(1-r\right)\right]}^2\left[2\beta \left(2-3r\right)-{\alpha}^2\left(1-r\right)\right]}{{\left[2\beta \left(2-r\right)-{\alpha}^2\left(1-r\right)\right]}^3}$; if r < (4β − α²)/(6β − α² ), then $\partial {\pi}_{mi}^{NN^{\ast }}/\partial \alpha >0$, if r > (4β − α²)/(6β − α² ), then $\partial {\pi}_{mi}^{NN^{\ast }}/\partial \alpha <0$.

Proof of Proposition 2

Similar to proof of Proposition 1, the reverse induction is applied here as well. In this set-up, we get p_i first which can be represented as $p_i^\ast=\left(a+\left(1-r\right)\left(\alpha\tau_i+w_i\right)\right)/\left(2\left(1-r\right)\right)$, i = 1, 2.

Then we substitute the p_i into Eq. (4), and we have the first partial conditions which are $\partial u_{m1}/\partial w_1=\frac{1}{2}\left[a\left(1+2\lambda+\eta\right)+c\left(1+\lambda+\eta+r\eta\right)+\alpha\tau_1-\alpha r \tau_2-2w_1 +rw_2+2 \alpha\tau_1\lambda-2\alpha r\tau_2\lambda-3w_1\lambda+2rw_2\lambda+\alpha\tau_1\eta-\alpha r\tau_2\eta-2w_1\eta\right]$, $\partial {u}_{m2}/\partial {w}_2=\frac{1}{2}\left[a\left(1+2\lambda +\eta \right)+c\left(1+\lambda +\eta + r\eta \right)+\alpha {\tau}_2-\alpha r{\tau}_1-2{w}_2+ r{w}_1+2\alpha{\tau}_2\lambda -2\alpha r{\tau}_1\lambda -3{w}_2\lambda +2r{w}_1\lambda +\alpha {\tau}_2\eta -\alpha r{\tau}_1\eta -2{w}_2\eta \right]$, $\partial {u}_{m1}/\partial {\tau}_1=\frac{1}{2}\left[\alpha \right({w}_1- a\lambda -\alpha {\tau}_1\lambda +\alpha r{\tau}_2\lambda +2{w}_1\lambda -r{w}_2 \lambda +{w}_1\eta +r{w}_2\eta -2\beta {\tau}_1\left(1+\lambda +\eta \right)- c\alpha \left(1+\lambda +\eta + r\eta \right)\Big]$, and $\partial {u}_{m2}/\partial {\tau}_2=\frac{1}{2}\left[\alpha \right({w}_2- a\lambda -\alpha {\tau}_2\lambda +\alpha r{\tau}_1\lambda +2{w}_2\lambda -r{w}_1 \lambda +{w}_2\eta +r{w}_1\eta -2\beta {\tau}_2\left(1+\lambda +\eta \right)- c\alpha \left(1+\lambda +\eta + r\eta \right)\Big]$. The second partial derivative $\partial^2u_{mi}/\partial w_i^2=-1-\eta-\frac{3}{2}\lambda$, $\partial^2u_{mi}/\partial\tau_i^2=-\frac{\alpha^2\lambda}{2}-\beta\left(1+\lambda+\eta\right)$, and $\partial^2u_{mi}/\partial w_i\partial\tau_i=\frac{\alpha}{2}\left(1+2\lambda+\eta\right)$; hence, the Hessian matrix can be shown as $H=\left(\begin{array}{c}-1-\eta -\frac{3}{2}\lambda \\ {}\frac{\alpha }{2}\left(1+2\lambda +\eta \right)\end{array}\right.\left.\begin{array}{c}\frac{\alpha }{2}\left(1+2\lambda +\eta \right)\\ {}-\frac{\alpha^2\lambda }{2}-\beta\left(1+\lambda +\eta \right)\end{array}\right)$, which is non-negative under condition β > (α²(1 + λ + η))/(2(2 + 3λ + 2η)). Then the u_mi is jointly concave in w_i and τ_i. Equating the first partial derivative to 0, we can get ${\tau}_i^{FF^{\ast }}=\frac{\alpha \left[a-c\left(1-r\right)\right]{\varDelta}_2}{\varDelta_3+{\varDelta}_1{\varDelta}_2}$ and ${w}_i^{FF^{\ast }}=\frac{2 a\beta \left(1+2\lambda +\eta \right)+c\left(2\beta +{\varDelta}_1\right){\varDelta}_2}{\varDelta_3+{\varDelta}_1{\varDelta}_2}$. Replacing w_i and τ_i with ${w}_i^{FF^{\ast }}$ and ${\tau}_i^{FF^{\ast }}$ into p_i^* and ${p}_i^{FF^{\ast }}$ can be obtained as ${p}_i^{FF^{\ast }}=\frac{c\left(\beta +{\varDelta}_1\right)\left(1-r\right){\varDelta}_2+ a\beta \left[5\lambda +3\left(1+\eta \right)-r\left(2+4\lambda +\eta \right)\right]}{\left(1-r\right)\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}$, where Δ₁ = α²(r − 1), Δ₂ = 1 + λ + η + rη, and Δ₃ = 2β(2 − r + 3λ − 2rλ + 2η). Finally, substituting ${w}_i^{FF^{\ast }}$, ${\tau}_i^{FF^{\ast }}$, and ${p}_i^{FF^{\ast }}$ into Eqs. (1)–(4), we can get ${\pi}_r^{FF^{\ast }}$, ${\pi}_{mi}^{FF^{\ast }}$, ${\pi}^{FF^{\ast }}$, and ${u}_{mi}^{FF^{\ast }}$.

Proof of Corollary 3

Similar to proof of Corollary 1, taking partial derivative of ${\tau}_i^{FF^{\ast }}$, ${w}_i^{FF^{\ast }}$, and ${p}_i^{FF^{\ast }}$ with respect to α, we can get

$\frac{\partial {\tau}_i^{FF^{\ast }}}{\partial \alpha }=\frac{\left(a-c+ cr\right){\varDelta}_2\left({\varDelta}_3-{\varDelta}_1{\varDelta}_2\right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}>0,$ $\frac{\partial {w}_i^{FF^{\ast }}}{\partial \alpha }=\frac{4\alpha \beta \left(a-c+ cr\right)\left(1-r\right)\left(1+2\lambda +\eta \right){\varDelta}_2}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}>0$, and $\frac{\partial {p}_i^{FF^{\ast }}}{\partial \alpha }=\frac{2\alpha \beta \left(a-c+ cr\right)\left[5{\lambda}^2+8\lambda \left(1+\eta \right)+3{\left(1+\eta \right)}^2+{r}^2\eta \left(2+4\lambda +\eta \right)+r\left(2+6\lambda +4{\lambda}^2-2{\eta}^2\right)\right]}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}>0$.

Proof of Corollary 4

Similar to proof of Corollary 3, taking partial derivative of ${D}_i^{FF^{\ast }}$, ${\pi}_r^{FF^{\ast }}$, ${\pi}_{mi}^{FF^{\ast }}$, and ${u}_{mi}^{FF^{\ast }}$ with respect to α, the results can be obtained. Thus, the process is omitted here.

Proof of Corollary 5

Based on the equilibrium solutions obtained in Proposition 2, taking partial derivative of ${\tau}_i^{FF^{\ast }}$, ${w}_i^{FF^{\ast }}$, and ${p}_i^{FF^{\ast }}$ with respect to fairness concern coefficients λ and η, respectively, we can get

• (1) $\frac{\partial {\tau}_i^{FF^{\ast }}}{\partial \lambda }=\frac{2\alpha \beta \left(a-c+ cr\right)\left(r-1\right)\left(1+\eta +2 r\eta \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}<0$, $\frac{\partial {\tau}_i^{FF^{\ast }}}{\partial \eta }=\frac{2\alpha \beta \left(a-c+ cr\right)\left(1-r\right)\left(r+\lambda +2 r\lambda \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}>0$.

• (2) $\frac{\partial {w}_i^{FF^{\ast }}}{\partial \lambda }=\frac{2\beta \left(a-c+ cr\right)\left(2\beta -{\alpha}^2+{\alpha}^2r\right)\left(1+\eta +2 r\eta \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}$, $\frac{\partial {w}_i^{FF^{\ast }}}{\partial \eta }=-\frac{2\beta \left(a-c+ cr\right)\left(2\beta -{\alpha}^2+{\alpha}^2r\right)\left(r+\lambda +2 r\lambda \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}$.

Thus, when 2β − α² + α²r > 0, then $\partial {w}_i^{FF^{\ast }}/\partial \lambda >0$ and $\partial {w}_i^{FF^{\ast }}/\partial \eta <0$; otherwise, $\partial {w}_i^{FF^{\ast }}/\partial \lambda <0$ and $\partial {w}_i^{FF^{\ast }}/\partial \eta >0$.

• (3) $\frac{\partial {p}_i^{FF^{\ast }}}{\partial \lambda }=\frac{2\beta \left(a-c+ cr\right)\left(\beta -{\alpha}^2+{\alpha}^2r\right)\left(1+\eta +2 r\eta \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}$, $\frac{\partial {p}_i^{FF^{\ast }}}{\partial \eta }=-\frac{2\beta \left(a-c+ cr\right)\left(\beta -{\alpha}^2+{\alpha}^2r\right)\left(r+\lambda +2 r\lambda \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}$.

Thus, when β − α² + α²r > 0, then $\partial {p}_i^{FF^{\ast }}/\partial \lambda >0$ and $\partial {p}_i^{FF^{\ast }}/\partial \eta <0$; otherwise, $\partial {p}_i^{FF^{\ast }}/\partial \lambda <0$ and $\partial {p}_i^{FF^{\ast }}/\partial \eta >0$.

Proof of Corollary 6

The proof is similar to that of Corollary 5 and is thus omitted here.

Proof of Proposition 3

From Proposition 1 and Proposition 2, we have

$\varDelta {\tau}_i^{\ast }={\tau}_i^{FF^{\ast }}-{\tau}_i^{NN^{\ast }}=\frac{2\alpha \beta \left(a-c+ cr\right)\left(1-r\right)\left( r\eta -\lambda \right)}{\left(4\beta -2\beta r+{\varDelta}_1\right)\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}$, $\varDelta {w}_i^{\ast }={w}_i^{FF^{\ast }}-{w}_i^{NN^{\ast }}=\frac{2\beta \left(a-c+ cr\right)\left(2\beta -{\alpha}^2+{\alpha}^2r\right)\left(\lambda - r\eta \right)}{\left(4\beta -2\beta r+{\varDelta}_1\right)\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}$, and $\varDelta {p}_i^{\ast }={p}_i^{FF^{\ast }}-{p}_i^{NN^{\ast }}=\frac{2\beta \left(a-c+ cr\right)\left(\beta -{\alpha}^2+{\alpha}^2r\right)\left(\lambda - r\eta \right)}{\left(4\beta -2\beta r+{\varDelta}_1\right)\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}$.

• (1) If λ < η and 0 < r < λ/η, then $\varDelta {\tau}_i^{\ast }>0$, namely ${\tau}_i^{FF^{\ast }}>{\tau}_i^{NN^{\ast }}$; when $\alpha <\sqrt{2\beta /\left(1-r\right)}$, then ${w}_i^{FF^{\ast }}>{w}_i^{NN^{\ast }}$; when $\alpha <\sqrt{\beta /\left(1-r\right)}$, then ${p}_i^{FF^{\ast }}>{p}_i^{NN^{\ast }}$; else if λ/η < r < 1, we have ${w}_i^{FF^{\ast }}<{w}_i^{NN^{\ast }}$, ${p}_i^{FF^{\ast }}<{p}_i^{NN^{\ast }}$, and ${\tau}_i^{FF^{\ast }}>{\tau}_i^{NN^{\ast }}$.

• (2) If λ > η, then due to Assumption 4, we have 0 < r < 1; thus, rη < λ; similarly, we can derive ${w}_i^{FF^{\ast }}>{w}_i^{NN^{\ast }}$, ${p}_i^{FF^{\ast }}>{p}_i^{NN^{\ast }}$, and ${\tau}_i^{FF^{\ast }}<{\tau}_i^{NN^{\ast }}$.

Proof of Corollary 7

According to the Proposition 4, we derive

• (1) $\frac{\partial \varDelta {\tau}_i^{\ast }}{\partial \lambda }=\frac{2\alpha \beta \left(a-c+ cr\right)\left(r-1\right)\left(1+\eta +2 r\eta \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}<0$; $\frac{\partial \varDelta {\tau}_i^{\ast }}{\partial \eta }=\frac{2\alpha \beta \left(a-c+ cr\right)\left(r-1\right)\left(r+\lambda +2 r\lambda \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}>0$.

• (2) $\frac{\partial \varDelta {w}_i^{\ast }}{\partial \lambda }=\frac{2\beta \left(a-c+ cr\right)\left(2\beta -{\alpha}^2+{\alpha}^2r\right)\left(1+\eta +2 r\eta \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}$, $\frac{\partial \varDelta {w}_i^{\ast }}{\partial \eta }=-\frac{2\beta \left(a-c+ cr\right)\left(2\beta -{\alpha}^2+{\alpha}^2r\right)\left(r+\lambda +2 r\lambda \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}$.

Thus, if $\alpha <\sqrt{2\beta /\left(1-r\right)}$, then $\partial \varDelta {w}_i^{\ast }/\partial \lambda >0$ and $\partial \varDelta {w}_i^{\ast }/\partial \eta <0$; if $\alpha >\sqrt{2\beta /\left(1-r\right)}$, then $\partial \varDelta {w}_i^{\ast }/\partial \lambda <0$ and $\partial \varDelta {w}_i^{\ast }/\partial \eta >0$.

• (3) $\frac{\partial \varDelta {p}_i^{\ast }}{\partial \lambda }=\frac{2\beta \left(a-c+ cr\right)\left(\beta -{\alpha}^2+{\alpha}^2r\right)\left(1+\eta +2 r\eta \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}$, $\frac{\partial \varDelta {p}_i^{\ast }}{\partial \eta }=-\frac{2\beta \left(a-c+ cr\right)\left(\beta -{\alpha}^2+{\alpha}^2r\right)\left(r+\lambda +2 r\lambda \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}$.

Thus, if $\alpha <\sqrt{\beta /\left(1-r\right)}$, then $\partial \varDelta {p}_i^{\ast }/\partial \lambda >0$ and $\partial \varDelta {p}_i^{\ast }/\partial \eta <0$; if $\alpha >\sqrt{\beta /\left(1-r\right)}$, then $\partial \varDelta {p}_i^{\ast }/\partial \lambda <0$ and $\partial \varDelta {p}_i^{\ast }/\partial \eta >0$.

Proof of Proposition 4

$\varDelta {\pi}_r^{\ast }={\pi}_r^{FF^{\ast }}-{\pi}_r^{NN^{\ast }}=\frac{8{\beta}^3{\left(a-c+ cr\right)}^2\left[\beta \left(4+5\lambda -2r-3 r\lambda + r\eta +4\eta -{r}^2\eta \right)+{\varDelta}_1{\varDelta}_2\right]\left( r\eta -\lambda \right)}{{\left(4\beta -2\beta r+{\varDelta}_1\right)}^2{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^2}$, as we know Δ₃ + Δ₁Δ₂ > 0; hence, when rη − λ > 0 is satisfied, then $\varDelta {\pi}_r^{\ast }>0$, namely, ${\pi}_r^{FF^{\ast }}>{\pi}_r^{NN^{\ast }}$ can be obtained.

Proof of Corollary 8

According to Proposition 5, we derive

$\frac{\partial \varDelta {\pi}_r^{\ast }}{\partial \lambda }=-\frac{8{\beta}^3{\left(a-c+ cr\right)}^2{\varDelta}_2\left(1+\eta +2 r\eta \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^3}<0$and $\frac{\partial \varDelta {\pi}_r^{\ast }}{\partial \eta }=\frac{8{\beta}^3{\left(a-c+ cr\right)}^2{\varDelta}_2\left(r+\lambda +2 r\lambda \right)}{{\left({\varDelta}_3+{\varDelta}_1{\varDelta}_2\right)}^3}>0$.