An artificial proof of a geometric inequality in a triangle

Abstract

In this paper, the authors give an artificial proof of a geometric inequality relating to the medians and the exradius in a triangle by making use of certain analytical techniques for systems of nonlinear algebraic equations.

MSC:51M16, 52A40.

1 Introduction and main results

For a given $\mathrm{△}ABC$, let a, b and c denote the side-lengths facing the angles A, B and C, respectively. Also, let $m_a$, $m_b$ and $m_c$ denote the corresponding medians, $r_a$, $r_b$ and $r_c$ the corresponding exradii, $s=\frac{1}{2}(a+b+c)$ the semi-perimeter, Δ the area. In addition, we let

$$\begin{array}{c}{m}_1=\frac{1}{2}\sqrt{{(b+c)}^2-a^2}=\sqrt{s(s-a)},\hfill \\ m_2=\frac{1}{2}\sqrt{2a^2+\frac{1}{4}{(b+c)}^2},\hfill \end{array}$$

and

$$r_1=\frac{a\sqrt{s(s-a)}}{2(s-a)}.$$

Throughout this paper, we will customarily use the cyclic sum symbols as follows:

$$\sum f(a)=f(a)+f(b)+f(c)$$

and

$$\sum f(b,c)=f(a,b)+f(b,c)+f(c,a).$$

In 2003, Liu [1] found the following interesting geometric inequality relating to the medians and the exradius in a triangle with the computer software BOTTEMA invented by Yang [2–5], and Liu thought this inequality cannot be proved by a human.

Theorem 1.1 In $\mathrm{△}ABC$, the best constant k for the following inequality

$$\sum {(r_b-r_c)}^2\ge k\cdot \sum {(m_b-m_c)}^2$$

(1.1)

is the real root on the interval $(3,4)$ of the following equation

$$6\text{,}561k^4-14\text{,}256k^3-18\text{,}080k^2-25\text{,}344k+20\text{,}736=0.$$

(1.2)

Furthermore, the constant k has its numerical approximation given by 3.2817755127.

In this paper, the authors give an artificial proof of Theorem 1.1.

2 Preliminary results

In order to prove Theorem 1.1, we require the following results.

Lemma 2.1 In $\mathrm{△}ABC$, if $a\le b\le c$, then

$$r_a^2+r_b^2+r_c^2-(r_1^2+2m_1^2)\ge \frac{3s(s-a){(b-c)}^2}{4(s-b)(s-c)}.$$

(2.1)

Proof From $a=(s-b)+(s-c)$ and the formulas of the exradius $r_a=\frac{\mathrm{\Delta }}{s-a}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s-a}$, etc., we get

$$\begin{array}{r}{r}_a^2+r_b^2+r_c^2-(r_1^2+2m_1^2)\\ =[\frac{1}{{(s-a)}^2}+\frac{1}{{(s-b)}^2}+\frac{1}{{(s-c)}^2}]s(s-a)(s-b)(s-c)-\frac{a^{2}s(s-a)}{4{(s-a)}^2}-2s(s-a)\\ =\frac{1}{4}s(s-a)[\frac{4(s-b)(s-c)}{{(s-a)}^2}+\frac{4(s-b)(s-c)}{{(s-b)}^2}+\frac{4(s-b)(s-c)}{{(s-c)}^2}-\frac{a^2}{{(s-a)}^2}-8]\\ =\frac{1}{4}s(s-a)[\frac{4(s-b)(s-c)-a^2}{{(s-a)}^2}+4(\frac{s-c}{s-b}+\frac{s-b}{s-c}-2)]\\ =\frac{1}{4}s(s-a)[-\frac{{(b-c)}^2}{{(s-a)}^2}+\frac{4{(b-c)}^2}{(s-b)(s-c)}]\\ =\frac{1}{4}s(s-a){(b-c)}^2[\frac{4}{(s-b)(s-c)}-\frac{1}{{(s-a)}^2}].\end{array}$$

(2.2)

For $a\le b\le c$, we have

$$s-a\ge s-b\ge s-c>0,$$

then

$$0<\frac{1}{s-a}\le \frac{1}{s-b}\le \frac{1}{s-c},$$

hence

$$\frac{1}{(s-b)(s-c)}\ge \frac{1}{{(s-a)}^2}>0.$$

(2.3)

Inequality (2.1) follows from inequalities (2.2)-(2.3) immediately. □

Lemma 2.2 In $\mathrm{△}ABC$, we have

$$(m_b+m_2)(m_c+m_2)\ge 4s\sqrt{(s-b)(s-c)}$$

(2.4)

and

$$a{(m_b+m_c)}^2-8s(s-b)(s-c)\ge \frac{3s\sqrt{(s-b)(s-c)}{(b-c)}^2}{a}.$$

(2.5)

Proof of inequality (2.4) From

$$m_2^2-\frac{1}{2}as=\frac{1}{4}{(a-\frac{b+c}{2})}^2\ge 0,$$

we immediately obtain

$$m_2\ge \sqrt{\frac{1}{2}as}.$$

(2.6)

In view of the AM-GM inequality, we get

$$\frac{a}{2}=\frac{(s-b)+(s-c)}{2}\ge \sqrt{(s-b)(s-c)}.$$

(2.7)

By the power mean inequality, we have

$$\sqrt{\frac{a}{2}}=\sqrt{\frac{(s-b)+(s-c)}{2}}\ge \frac{\sqrt{s-b}+\sqrt{s-c}}{2}.$$

(2.8)

By the well-known inequalities $m_b\ge \sqrt{s(s-b)}$ and $m_c\ge \sqrt{s(s-c)}$, together with inequalities (2.6)-(2.8), we obtain

$$\begin{array}{c}(m_b+m_2)(m_c+m_2)\hfill \\ \ge (\sqrt{s(s-b)}+\sqrt{\frac{1}{2}as})(\sqrt{s(s-c)}+\sqrt{\frac{1}{2}as})\hfill \\ =s(\sqrt{s-b}+\sqrt{\frac{1}{2}a})(\sqrt{s-c}+\sqrt{\frac{1}{2}a})\hfill \\ =s[\frac{1}{2}a+\sqrt{\frac{1}{2}a}(\sqrt{s-b}+\sqrt{s-c})+\sqrt{(s-b)(s-c)}]\hfill \\ \ge s[\frac{1}{2}{(\sqrt{s-b}+\sqrt{s-c})}^2+2\sqrt{(s-b)(s-c)}]\hfill \\ =s[\frac{1}{2}a+3\sqrt{(s-b)(s-c)}]\hfill \\ \ge 4s\sqrt{(s-b)(s-c)}.\hfill \end{array}$$

The proof of inequality (2.4) is thus complete. □

Proof of inequality (2.5) According to the well-known inequalities $m_b\ge \sqrt{s(s-b)}$, $m_c\ge \sqrt{s(s-c)}$ and inequality (2.7), we have

$$\begin{array}{l}a{(m_b+m_c)}^2-8s(s-b)(s-c)\\ =[a-2\sqrt{(s-b)(s-c)}]{(m_b+m_c)}^2\\ +2\sqrt{(s-b)(s-c)}[{(m_b+m_c)}^2-4s\sqrt{(s-b)(s-c)}]\\ \ge [a-2\sqrt{(s-b)(s-c)}]\cdot 4m_b{m}_c+2\sqrt{(s-b)(s-c)}[{(\sqrt{s(s-b)}+\sqrt{s(s-c)})}^2\\ -4s\sqrt{(s-b)(s-c)}]\\ \ge 4s[a-2\sqrt{(s-b)(s-c)}]\sqrt{(s-b)(s-c)}+2\sqrt{(s-b)(s-c)}[a-2\sqrt{(s-b)(s-c)}]\\ =6s\sqrt{(s-b)(s-c)}[a-2\sqrt{(s-b)(s-c)}]\\ =\frac{6s\sqrt{(s-b)(s-c)}{(b-c)}^2}{a+2\sqrt{(s-b)(s-c)}}\\ \ge \frac{3s\sqrt{(s-b)(s-c)}{(b-c)}^2}{a}.\end{array}$$

(2.9)

Hence, we complete the proof of inequality (2.5). □

Lemma 2.3 In $\mathrm{△}ABC$, we have

$$m_b{m}_c\le m_2^2.$$

(2.10)

Proof From the formulas of the medians, we have

$$\begin{array}{rcl}{m}_b{m}_c-m_2^2& =& \frac{m_b^2{m}_c^2-m_2^4}{m_b{m}_c+m_2^2}\\ =& \frac{\frac{1}{16}(2c^2+2a^2-b^2)(2a^2+2b^2-c^2)-\frac{1}{16}(2a^2+\frac{1}{4}{(b+c)}^2)}{m_b{m}_c+m_2^2}\\ =& \frac{\{16[a^2-{(b+c)}^2]-(17b^2+17c^2+38bc)\}{(b-c)}^2}{256(m_b{m}_c+m_2^2)}\le 0.\end{array}$$

Therefore, inequality (2.10) holds true. □

Lemma 2.4 In $\mathrm{△}ABC$, if $a\le b\le c$, then

$$\begin{array}{r}\frac{m_b+m_c}{m_a+m_1}+\frac{1}{4}(\frac{m_1}{m_2+m_b}+\frac{m_1}{m_2+m_c})-\frac{9{(b+c)}^2}{8{(m_b+m_c)}^2}\\ \ge \frac{m_2}{m_1}+\frac{m_1}{4m_2}-\frac{9a{(b+c)}^2}{64s(s-b)(s-c)}.\end{array}$$

(2.11)

Proof It is obvious that $m_b>c-\frac{b}{2}$ and $m_c>b-\frac{c}{2}$, then we have $m_b+m_c>\frac{1}{2}(b+c)$, thus

$${(m_b-m_c)}^2=\frac{{(m_b^2-m_c^2)}^2}{{(m_b+m_c)}^2}=\frac{9{(b+c)}^2{(b-c)}^2}{16{(m_b+m_c)}^2}\le \frac{9}{4}{(b-c)}^2.$$

(2.12)

For $a\le b\le c$, we have that

$$m_a\ge \{\begin{array}{c}{m}_1\hfill \\ m_b\hfill \end{array}\ge m_2\ge m_c.$$

(2.13)

By Lemma 2.3 and inequalities (2.12)-(2.13), we have

$$\begin{array}{l}\frac{m_b+m_c}{m_a+m_1}+\frac{1}{4}(\frac{m_1}{m_2+m_b}+\frac{m_1}{m_2+m_c})-\frac{m_2}{m_1}-\frac{m_1}{4m_2}\\ =\frac{m_b+m_c-2m_2}{m_a+m_1}+\frac{m_2(m_1-m_a)}{m_1(m_a+m_1)}+\frac{m_1(m_2^2-m_b{m}_c)}{4m_2(m_2+m_b)(m_2+m_c)}\\ \ge \frac{{(m_b+m_c)}^2-4m_2^2}{(m_a+m_1)(m_b+m_c+2m_2)}+\frac{m_2(m_1^2-m_a^2)}{m_1{(m_a+m_1)}^2}\\ =\frac{2(m_b^2+m_c^2)-{(m_b-m_c)}^2-4m_2^2}{(m_a+m_1)(m_b+m_c+2m_2)}+\frac{m_2(m_1^2-m_a^2)}{m_1{(m_a+m_1)}^2}\\ =\frac{\frac{1}{4}{(b-c)}^2-{(m_b-m_c)}^2}{(m_a+m_1)(m_b+m_c+2m_2)}-\frac{m_2{(b-c)}^2}{4m_1{(m_a+m_1)}^2}\\ \ge \frac{\frac{1}{4}{(b-c)}^2-\frac{9}{4}{(b-c)}^2}{(m_a+m_1)(m_b+m_c+2m_2)}-\frac{{(b-c)}^2}{4{(m_a+m_1)}^2}\\ =\frac{-2{(b-c)}^2}{(m_a+m_1)(m_b+m_c+2m_2)}-\frac{{(b-c)}^2}{4{(m_a+m_1)}^2}\\ \ge \frac{-2{(b-c)}^2}{(m_a+m_1)(m_b+m_c)}-\frac{{(b-c)}^2}{4{(m_a+m_1)}^2}\\ =\frac{-2{(b-c)}^2}{{(m_b+m_c)}^2}-\frac{{(b-c)}^2}{4{(m_b+m_c)}^2}\\ \ge \frac{-9{(b-c)}^2}{4{(m_b+m_c)}^2}.\end{array}$$

(2.14)

By inequality (2.5), (2.7) and $a\le b\le c$, we obtain that

$$\begin{array}{l}\frac{9a{(b+c)}^2}{64s(s-b)(s-c)}-\frac{9{(b+c)}^2}{8{(m_b+m_c)}^2}\\ =\frac{9{(b+c)}^2[a{(m_b+m_c)}^2-8s(s-b)(s-c)]}{64s(s-b)(s-c){(m_b+m_c)}^2}\\ \ge \frac{9{(b+c)}^2}{64s(s-b)(s-c){(m_b+m_c)}^2}\cdot \frac{3s\sqrt{(s-b)(s-c)}{(b-c)}^2}{a}\\ =\frac{27{(b+c)}^2{(b-c)}^2}{64a\sqrt{(s-b)(s-c)}{(m_b+m_c)}^2}\\ \ge \frac{27{(b+c)}^2{(b-c)}^2}{32a^2{(m_b+m_c)}^2}\\ \ge \frac{27{(b-c)}^2}{8{(m_b+m_c)}^2}.\end{array}$$

(2.15)

By inequalities (2.14)-(2.15), we have

$$\begin{array}{l}[\frac{m_b+m_c}{m_a+m_1}+\frac{1}{4}(\frac{m_1}{m_2+m_b}+\frac{m_1}{m_2+m_c})-\frac{9{(b+c)}^2}{8{(m_b+m_c)}^2}]\\ -[\frac{m_2}{m_1}+\frac{m_1}{4m_2}-\frac{9a{(b+c)}^2}{64s(s-b)(s-c)}]\\ =[\frac{m_b+m_c}{m_a+m_1}+\frac{1}{4}(\frac{m_1}{m_2+m_b}+\frac{m_1}{m_2+m_c})-\frac{m_2}{m_1}-\frac{m_1}{4m_2}]\\ +[\frac{9a{(b+c)}^2}{64s(s-b)(s-c)}-\frac{9{(b+c)}^2}{8{(m_b+m_c)}^2}]\\ \ge \frac{-9{(b-c)}^2}{4{(m_b+m_c)}^2}+\frac{27{(b-c)}^2}{8{(m_b+m_c)}^2}\\ =\frac{9{(b-c)}^2}{8{(m_b+m_c)}^2}\ge 0.\end{array}$$

(2.16)

Inequality (2.11) follows from inequality (2.16) immediately. □

Lemma 2.5 In $\mathrm{△}ABC$, if $a\le b\le c$, then

$$\frac{m_2}{m_1}+\frac{m_1}{4m_2}+\frac{3{(b+c)}^2}{16a^2}\ge 2$$

(2.17)

and

$$\frac{m_1+\sqrt{3}a}{s}\le \sqrt{3}.$$

(2.18)

Proof Without loss of generality, we can take $b+c=2$ and $a=x$, for $a\le b\le c$, we have $0<x\le 1$.

• (i) First, we prove inequality (2.17).

  $$\begin{array}{rl}\frac{m_2}{m_1}+\frac{m_1}{4m_2}+\frac{3{(b+c)}^2}{16a^2}-2& =\sqrt{\frac{1+2x^2}{4-x^2}}+\frac{1}{4}\sqrt{\frac{4-x^2}{1+2x^2}}+\frac{3}{4x^2}-2\\ =\frac{8+7x^2}{4\sqrt{(4-x^2)(1+2x^2)}}+\frac{3(1-x^2)}{4x^2}-\frac{5}{4}\\ \ge \frac{8+7x^2}{4\cdot \frac{(4-x^2)+(1+2x^2)}{2}}+\frac{3(1-x^2)}{4x^2}-\frac{5}{4}\\ =\frac{8+7x^2}{2(5+x^2)}+\frac{3(1-x^2)}{4x^2}-\frac{5}{4}\\ =\frac{9(x^2-1)}{4(5+x^2)}+\frac{3(1-x^2)}{4x^2}\\ \ge \frac{3(x^2-1)}{8}+\frac{3(1-x^2)}{4}\\ =\frac{3(1-x^2)}{8}\ge 0.\end{array}$$

  (2.19)

Inequality (2.19) terminates the proof of inequality (2.17).

• (ii) Second, we prove inequality (2.18).

  $$\begin{array}{l}{m}_1+\sqrt{3}a-\sqrt{3}s\\ =\frac{1}{2}\sqrt{4-x^2}-\frac{\sqrt{3}}{2}(2-x)\\ =\frac{1}{2}\sqrt{2-x}(\sqrt{2+x}-\sqrt{3(2-x)})\\ =\frac{-2\sqrt{2-x}(1-x)}{\sqrt{2+x}+\sqrt{3(2-x)}}\le 0.\end{array}$$

  (2.20)

Inequality (2.18) follows from inequality (2.20) immediately. □

Lemma 2.6 In $\mathrm{△}ABC$, if $a\le b\le c$, then

$$m_a{m}_b+m_b{m}_c+m_c{m}_a-2m_1{m}_2-m_2^2\ge \frac{3}{8}{(b-c)}^2-\frac{3s(s-a){(b-c)}^2}{16(s-b)(s-c)}.$$

(2.21)

Proof By the AM-GM inequality, the well-known inequalities $m_b\ge \sqrt{s(s-b)}$ and $m_c\ge \sqrt{s(s-c)}$, we get

$${(m_b+m_c)}^2\ge 4m_b{m}_c\ge 4s\sqrt{(s-b)(s-c)}\ge 6a\sqrt{(s-b)(s-c)}\ge 12(s-b)(s-c)$$

or

$$m_b+m_c\ge 2\sqrt{3}\sqrt{(s-b)(s-c)}.$$

(2.22)

By inequalities (2.4), (2.10), (2.11), (2.17), (2.22), we obtain that

$$\begin{array}{c}{m}_a{m}_b+m_b{m}_c+m_c{m}_a-2m_1{m}_2-m_2^2\hfill \\ =\frac{(m_b+m_c)(m_a^2-m_1^2)}{m_a+m_1}+\frac{m_1(m_b^2-m_2^2)}{m_b+m_2}+\frac{m_1(m_c^2-m_2^2)}{m_c+m_2}-\frac{{(m_b^2-m_c^2)}^2}{2{(m_b+m_c)}^2}+\frac{1}{16}{(b-c)}^2\hfill \\ =\frac{(m_b+m_c){(b-c)}^2}{4(m_a+m_1)}+\frac{m_1(5b+7c)(c-b)}{16(m_b+m_2)}+\frac{m_1(7b+5c)(b-c)}{16(m_c+m_2)}\hfill \\ -\frac{9{(b+c)}^2{(b-c)}^2}{32{(m_b+m_c)}^2}+\frac{1}{16}{(b-c)}^2\hfill \\ =\frac{(m_b+m_c){(b-c)}^2}{4(m_a+m_1)}+\frac{m_1{(b-c)}^2}{16(m_b+m_2)}+\frac{m_1{(b-c)}^2}{16(m_c+m_2)}\hfill \\ -\frac{9m_1{(b+c)}^2{(b-c)}^2}{32(m_b+m_2)(m_c+m_2)(m_b+m_c)}\hfill \\ -\frac{9{(b+c)}^2{(b-c)}^2}{32{(m_b+m_c)}^2}+\frac{1}{16}{(b-c)}^2\hfill \\ \ge \frac{1}{4}(\frac{m_2}{m_1}+\frac{m_1}{4m_2}-\frac{9a{(b+c)}^2}{64s(s-b)(s-c)}-\frac{m_1{(b+c)}^2}{64\sqrt{3}s(s-b)(s-c)}+\frac{1}{4}){(b-c)}^2\hfill \\ =\frac{1}{4}(\frac{m_2}{m_1}+\frac{m_1}{4m_2}-\frac{9(m_1+\sqrt{3}a){(b+c)}^2}{64\sqrt{3}s(s-b)(s-c)}+\frac{1}{4}){(b-c)}^2\hfill \\ \ge \frac{1}{4}(\frac{m_2}{m_1}+\frac{m_1}{4m_2}-\frac{9{(b+c)}^2}{64(s-b)(s-c)}+\frac{1}{4}){(b-c)}^2\hfill \\ =\frac{1}{4}(\frac{m_2}{m_1}+\frac{m_1}{4m_2}-\frac{3[{(b+c)}^2-a^2]}{16(s-b)(s-c)}+\frac{3[{(b+c)}^2-4a^2]}{64(s-b)(s-c)}+\frac{1}{4}){(b-c)}^2\hfill \\ \ge \frac{1}{4}(\frac{m_2}{m_1}+\frac{m_1}{4m_2}-\frac{3s(s-a)}{4(s-b)(s-c)}+\frac{3[{(b+c)}^2-4a^2]}{16a^2}+\frac{1}{4}){(b-c)}^2\hfill \\ =\frac{1}{4}(\frac{m_2}{m_1}+\frac{m_1}{4m_2}+\frac{3{(b+c)}^2}{16a^2}-\frac{3s(s-a)}{4(s-b)(s-c)}-\frac{1}{2}){(b-c)}^2\hfill \\ \ge \frac{1}{4}(2-\frac{3s(s-a)}{4(s-b)(s-c)}-\frac{1}{2}){(b-c)}^2\hfill \\ =\frac{3}{8}{(b-c)}^2-\frac{3s(s-a){(b-c)}^2}{16(s-b)(s-c)}.\hfill \end{array}$$

The proof of Lemma 2.6 is thus completed. □

Lemma 2.7 In $\mathrm{△}ABC$, if inequality (1.1) holds, then $k\le 4$.

Proof Let $b=c=1$ and $a=x$. For $a\le b\le c$, we have $x\in (0,1]$, then inequality (1.1) is equivalent to

$$\begin{array}{r}2{(\frac{x\sqrt{4-x^2}}{2(2-x)}-\frac{\sqrt{4-x^2}}{2})}^2\ge 2k{(\frac{\sqrt{4-x^2}}{2}-\frac{\sqrt{2x^2+1}}{2})}^2\\ ⟺\frac{2+x}{2-x}\ge k\cdot \frac{9{(1+x)}^2}{4{(\sqrt{4-x^2}+\sqrt{2x^2+1})}^2}\\ ⟺k\le \frac{4(2+x){(\sqrt{4-x^2}+\sqrt{2x^2+1})}^2}{9(2-x){(1+x)}^2}.\end{array}$$

(2.23)

Taking $x=1$ in inequality (2.23), we obtain that $k\le 4$. □

Lemma 2.8 In $\mathrm{△}ABC$, if $a\le b\le c$ and $0<k\le 4$, then we have

$$\sum {(r_b-r_c)}^2-k\cdot \sum {(m_b-m_c)}^2\ge 2{(r_1-m_1)}^2-2k{(m_1-m_2)}^2.$$

(2.24)

Proof For

$$\sum {(r_b-r_c)}^2=2\sum r_a^2-2\sum r_b{r}_c=2\sum r_a^2-2s^2$$

and

$$\sum {(m_b-m_c)}^2=2\sum m_a^2-2\sum m_b{m}_c=\frac{3}{2}\sum a^2-2\sum m_b{m}_c,$$

hence, by Lemmas 2.1 and 2.6, we have

$$\begin{array}{c}\sum {(r_b-r_c)}^2-k\cdot \sum {(m_b-m_c)}^2-2{(r_1-m_1)}^2+2k{(m_1-m_2)}^2\hfill \\ =2[\sum r_a^2-r_1^2-2m_1^2]+2k[\sum m_b{m}_c-2m_1{m}_2-m_2^2-\frac{3}{8}{(b-c)}^2]\hfill \\ \ge \frac{3s(s-a){(b-c)}^2}{2(s-b)(s-c)}-\frac{3ks(s-a){(b-c)}^2}{8(s-b)(s-c)}\hfill \\ =\frac{3(4-k)s(s-a){(b-c)}^2}{8(s-b)(s-c)}\ge 0.\hfill \end{array}$$

The proof of Lemma 2.8 is complete. □

Lemma 2.9 (see [4, 6, 7])

Define

$$F(x)=a_0{x}^n+a_1{x}^{n-1}+\cdots +a_n,$$

and

$$G(x)=b_0{x}^m+b_1{x}^{m-1}+\cdots +b_m.$$

If $a_0\ne 0$ or $b_0\ne 0$, then the polynomials $F(x)$ and $G(x)$ have a common root if and only if

where $R(F,G)$ ($(m+n)\times (m+n)$ determinant) is Sylvester’s resultant of $F(x)$ and $G(x)$.

Lemma 2.10 (see [7, 8])

Given a polynomial $f(x)$ with real coefficients

$$f(x)=a_0{x}^n+a_1{x}^{n-1}+\cdots +a_n,$$

if the number of the sign changes in the revised sign list of its discriminant sequence

$$\{D_1(f),D_2(f),\dots ,D_n(f)\}$$

is v, then the number of the pairs of distinct conjugate imaginary roots of $f(x)$ equals v. Furthermore, if the number of non-vanishing members in the revised sign list is l, then the number of the distinct real roots of $f(x)$ equals $l-2v$.

3 The proof of Theorem 1.1

Proof If $k\le 0$,we can easily find that inequality (1.1) holds. Hence, we only need to consider the case $k>0$, and by Lemma 2.7, we only need to consider the case $0<k\le 4$.

Now we determine the best constant k such that inequality (1.1) holds. Since inequality (1.1) is symmetrical with respect to the side-lengths a, b and c, there is no harm in supposing $a\le b\le c$. Thus, by Lemma 2.8, we only need to determine the best constant k such that

$$2{(r_1-m_1)}^2-2k{(m_1-m_2)}^2\ge 0$$

or, equivalently, that

$$\begin{array}{l}{(\frac{a\sqrt{{(b+c)}^2-a^2}}{2(b+c-a)}-\frac{\sqrt{{(b+c)}^2-a^2}}{2})}^2-k{(\frac{\sqrt{{(b+c)}^2-a^2}}{2}-\frac{1}{2}\sqrt{2a^2+\frac{1}{4}{(b+c)}^2})}^2\\ \ge 0.\end{array}$$

(3.1)

Without loss of generality, we can assume that

$$a=t\text{and}\frac{b+c}{2}=1(0<t\le 1),$$

because inequality (3.1) is homogeneous with respect to a and $\frac{b+c}{2}$. Thus, clearly, inequality (3.1) is equivalent to the following inequality:

$${(\frac{t\sqrt{4-t^2}}{2(2-t)}-\frac{\sqrt{4-t^2}}{2})}^2-k{(\frac{\sqrt{4-t^2}}{2}-\frac{\sqrt{2t^2+1}}{2})}^2\ge 0.$$

(3.2)

We consider the following two cases separately.

Case 1. When $t=1$, inequality (3.2) holds true for any $k\in R:=(-\mathrm{\infty },+\mathrm{\infty })$.

Case 2. When $0<t<1$, inequality (3.2) is equivalent to the following inequality:

$$k\le \frac{4(2+t){(\sqrt{4-t^2}+\sqrt{2t^2+1})}^2}{9(2-t){(1+t)}^2}.$$

(3.3)

Define the function

$$g(t):=\frac{4(2+t){(\sqrt{4-t^2}+\sqrt{2t^2+1})}^2}{9(2-t){(1+t)}^2},x\in (0,1).$$

Calculating the derivative for $g(t)$, we get

$$g^{\prime }(t)=\frac{8(\sqrt{4-t^2}+\sqrt{2t^2+1})\cdot \sqrt{4-t^2}\cdot [(2t^3+5t^2+10t-2)-(2-t)\sqrt{4-t^2}\cdot \sqrt{2t^2+1}]}{9{(2-t)}^2{(1+t)}^3\sqrt{2t^2+1}\cdot \sqrt{4-t^2}}.$$

By setting $g^{\prime }(t)=0$, we obtain

$$\sqrt{4-t^2}\cdot [(2t^3+5t^2+10t-2)-(2-t)\sqrt{4-t^2}\cdot \sqrt{2t^2+1}]=0.$$

(3.4)

It is easily observed that the equation $\sqrt{4-t^2}=0$ has no real root on the interval $(0,1)$. Hence, the roots of equation (3.4) are also solutions of the following equation:

$$(2t^3+5t^2+10t-2)-(2-t)\sqrt{4-t^2}\cdot \sqrt{2t^2+1}=0,$$

that is,

$${(1+t)}^2\phi (t)=0,$$

(3.5)

where

$$\phi (t)=t^4+10t^2-2.$$

It is obvious that the equation

$${(1+t)}^2=0$$

(3.6)

has no real root on the interval $(0,1)$.

It is easy to find that the equation

$$\phi (t)=0$$

(3.7)

has one positive real root. Moreover, it is not difficult to observe that $\phi (0)=-2<0$ and $\phi (1)=9>0$. We can thus find that equation (3.7) has one distinct real root on the interval $(0,1)$. So that equation (3.4) has only one real root $t_0$ given by $t_0=0.442890982868958\dots$ on the interval $(0,1)$, and

$$g{(t)}_{\max }=g(t_0)\approx 3.2817755127\in (3,4).$$

(3.8)

Now we prove $g(t_0)$ is the root of equation (1.2). For this purpose, we consider the following nonlinear algebraic equation system:

$$\{\begin{array}{c}\phi (t_0)=0,\hfill \\ 2t_0^2+1-u_0^2=0,\hfill \\ 4-t_0^2-v_0^2=0,\hfill \\ 4(2+t){(u_0+v_0)}^2-9(2-t){(1+t)}^{2}k=0.\hfill \end{array}$$

(3.9)

It is easy to see that $g(t_0)$ is also the solution of nonlinear algebraic equation system (3.9). If we eliminate the $v_0$, $u_0$ and $t_0$ ordinal by the resultant (by using Lemma 2.9), then we get

$$29\text{,}648\text{,}323\text{,}021\text{,}629\text{,}456\cdot {\varphi }_1^2(k)\cdot {\varphi }_2^2(k)=0,$$

(3.10)

where

$${\varphi }_1(k)=6\text{,}561k^4-14\text{,}256k^3-18\text{,}080k^2-25\text{,}344k+20\text{,}736$$

and

$${\varphi }_2(k)=729k^4-7\text{,}344k^3+8\text{,}800k^2-13\text{,}056k+2\text{,}304.$$

The revised sign list of the discriminant sequence of ${\varphi }_1(k)$ is given by

$$[1,1,-1,-1].$$

(3.11)

The revised sign list of the discriminant sequence of ${\varphi }_2(k)$ is given by

$$[1,1,-1,-1].$$

(3.12)

So the number of sign changes in the revised sign list of (3.11) and (3.12) are both 2. Thus, by applying Lemma 2.10, we find that the equations

$${\varphi }_1(k)=0$$

(3.13)

and

$${\varphi }_2(k)=0$$

(3.14)

both have two distinct real roots. In addition, it is easy to find that

$$\begin{array}{c}{\varphi }_1(0)=20\text{,}736>0;{\varphi }_2(0)=2\text{,}304>0,\hfill \\ {\varphi }_1(1)=-30\text{,}383<0;{\varphi }_2(1)=-8\text{,}567<0,\hfill \\ {\varphi }_1(3)=-71\text{,}487<0;{\varphi }_2(8)=-313\text{,}088<0\hfill \end{array}$$

and

$${\varphi }_1(4)=397\text{,}312>0;{\varphi }_2(9)=26\text{,}793>0.$$

We can thus find that equation (3.13) has two distinct real roots on the intervals

$$(0,1)\text{and}(3,4).$$

And equation (3.14) has two distinct real roots on the intervals

$$(0,1)\text{and}(8,9).$$

Hence, by (3.8), we can conclude that $g(t_0)$ is the root of equation (1.2). The proof of Theorem 1.1 is thus completed. □