# 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, $s=\frac{1}{2}\left(a+b+c\right)$ the semi-perimeter, Δ the area. In addition, we let

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

and

${r}_{1}=\frac{a\sqrt{s\left(s-a\right)}}{2\left(s-a\right)}.$

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

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

and

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

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 [25], 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

$\sum {\left({r}_{b}-{r}_{c}\right)}^{2}\ge k\cdot \sum {\left({m}_{b}-{m}_{c}\right)}^{2}$
(1.1)

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

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

${r}_{a}^{2}+{r}_{b}^{2}+{r}_{c}^{2}-\left({r}_{1}^{2}+2{m}_{1}^{2}\right)\ge \frac{3s\left(s-a\right){\left(b-c\right)}^{2}}{4\left(s-b\right)\left(s-c\right)}.$
(2.1)

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

$\begin{array}{r}{r}_{a}^{2}+{r}_{b}^{2}+{r}_{c}^{2}-\left({r}_{1}^{2}+2{m}_{1}^{2}\right)\\ \phantom{\rule{1em}{0ex}}=\left[\frac{1}{{\left(s-a\right)}^{2}}+\frac{1}{{\left(s-b\right)}^{2}}+\frac{1}{{\left(s-c\right)}^{2}}\right]s\left(s-a\right)\left(s-b\right)\left(s-c\right)-\frac{{a}^{2}s\left(s-a\right)}{4{\left(s-a\right)}^{2}}-2s\left(s-a\right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}s\left(s-a\right)\left[\frac{4\left(s-b\right)\left(s-c\right)}{{\left(s-a\right)}^{2}}+\frac{4\left(s-b\right)\left(s-c\right)}{{\left(s-b\right)}^{2}}+\frac{4\left(s-b\right)\left(s-c\right)}{{\left(s-c\right)}^{2}}-\frac{{a}^{2}}{{\left(s-a\right)}^{2}}-8\right]\\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}s\left(s-a\right)\left[\frac{4\left(s-b\right)\left(s-c\right)-{a}^{2}}{{\left(s-a\right)}^{2}}+4\left(\frac{s-c}{s-b}+\frac{s-b}{s-c}-2\right)\right]\\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}s\left(s-a\right)\left[-\frac{{\left(b-c\right)}^{2}}{{\left(s-a\right)}^{2}}+\frac{4{\left(b-c\right)}^{2}}{\left(s-b\right)\left(s-c\right)}\right]\\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}s\left(s-a\right){\left(b-c\right)}^{2}\left[\frac{4}{\left(s-b\right)\left(s-c\right)}-\frac{1}{{\left(s-a\right)}^{2}}\right].\end{array}$
(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

$\frac{1}{\left(s-b\right)\left(s-c\right)}\ge \frac{1}{{\left(s-a\right)}^{2}}>0.$
(2.3)

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

Lemma 2.2 In $\mathrm{△}ABC$, we have

$\left({m}_{b}+{m}_{2}\right)\left({m}_{c}+{m}_{2}\right)\ge 4s\sqrt{\left(s-b\right)\left(s-c\right)}$
(2.4)

and

$a{\left({m}_{b}+{m}_{c}\right)}^{2}-8s\left(s-b\right)\left(s-c\right)\ge \frac{3s\sqrt{\left(s-b\right)\left(s-c\right)}{\left(b-c\right)}^{2}}{a}.$
(2.5)

Proof of inequality (2.4) From

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

we immediately obtain

${m}_{2}\ge \sqrt{\frac{1}{2}as}.$
(2.6)

In view of the AM-GM inequality, we get

$\frac{a}{2}=\frac{\left(s-b\right)+\left(s-c\right)}{2}\ge \sqrt{\left(s-b\right)\left(s-c\right)}.$
(2.7)

By the power mean inequality, we have

$\sqrt{\frac{a}{2}}=\sqrt{\frac{\left(s-b\right)+\left(s-c\right)}{2}}\ge \frac{\sqrt{s-b}+\sqrt{s-c}}{2}.$
(2.8)

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

$\begin{array}{c}\left({m}_{b}+{m}_{2}\right)\left({m}_{c}+{m}_{2}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge \left(\sqrt{s\left(s-b\right)}+\sqrt{\frac{1}{2}as}\right)\left(\sqrt{s\left(s-c\right)}+\sqrt{\frac{1}{2}as}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=s\left(\sqrt{s-b}+\sqrt{\frac{1}{2}a}\right)\left(\sqrt{s-c}+\sqrt{\frac{1}{2}a}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=s\left[\frac{1}{2}a+\sqrt{\frac{1}{2}a}\left(\sqrt{s-b}+\sqrt{s-c}\right)+\sqrt{\left(s-b\right)\left(s-c\right)}\right]\hfill \\ \phantom{\rule{1em}{0ex}}\ge s\left[\frac{1}{2}{\left(\sqrt{s-b}+\sqrt{s-c}\right)}^{2}+2\sqrt{\left(s-b\right)\left(s-c\right)}\right]\hfill \\ \phantom{\rule{1em}{0ex}}=s\left[\frac{1}{2}a+3\sqrt{\left(s-b\right)\left(s-c\right)}\right]\hfill \\ \phantom{\rule{1em}{0ex}}\ge 4s\sqrt{\left(s-b\right)\left(s-c\right)}.\hfill \end{array}$

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

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

$\begin{array}{l}a{\left({m}_{b}+{m}_{c}\right)}^{2}-8s\left(s-b\right)\left(s-c\right)\\ \phantom{\rule{1em}{0ex}}=\left[a-2\sqrt{\left(s-b\right)\left(s-c\right)}\right]{\left({m}_{b}+{m}_{c}\right)}^{2}\\ \phantom{\rule{2em}{0ex}}+2\sqrt{\left(s-b\right)\left(s-c\right)}\left[{\left({m}_{b}+{m}_{c}\right)}^{2}-4s\sqrt{\left(s-b\right)\left(s-c\right)}\right]\\ \phantom{\rule{1em}{0ex}}\ge \left[a-2\sqrt{\left(s-b\right)\left(s-c\right)}\right]\cdot 4{m}_{b}{m}_{c}+2\sqrt{\left(s-b\right)\left(s-c\right)}\left[{\left(\sqrt{s\left(s-b\right)}+\sqrt{s\left(s-c\right)}\right)}^{2}\\ \phantom{\rule{2em}{0ex}}-4s\sqrt{\left(s-b\right)\left(s-c\right)}\right]\\ \phantom{\rule{1em}{0ex}}\ge 4s\left[a-2\sqrt{\left(s-b\right)\left(s-c\right)}\right]\sqrt{\left(s-b\right)\left(s-c\right)}+2\sqrt{\left(s-b\right)\left(s-c\right)}\left[a-2\sqrt{\left(s-b\right)\left(s-c\right)}\right]\\ \phantom{\rule{1em}{0ex}}=6s\sqrt{\left(s-b\right)\left(s-c\right)}\left[a-2\sqrt{\left(s-b\right)\left(s-c\right)}\right]\\ \phantom{\rule{1em}{0ex}}=\frac{6s\sqrt{\left(s-b\right)\left(s-c\right)}{\left(b-c\right)}^{2}}{a+2\sqrt{\left(s-b\right)\left(s-c\right)}}\\ \phantom{\rule{1em}{0ex}}\ge \frac{3s\sqrt{\left(s-b\right)\left(s-c\right)}{\left(b-c\right)}^{2}}{a}.\end{array}$
(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

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

Therefore, inequality (2.10) holds true. □

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

$\begin{array}{r}\frac{{m}_{b}+{m}_{c}}{{m}_{a}+{m}_{1}}+\frac{1}{4}\left(\frac{{m}_{1}}{{m}_{2}+{m}_{b}}+\frac{{m}_{1}}{{m}_{2}+{m}_{c}}\right)-\frac{9{\left(b+c\right)}^{2}}{8{\left({m}_{b}+{m}_{c}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}\ge \frac{{m}_{2}}{{m}_{1}}+\frac{{m}_{1}}{4{m}_{2}}-\frac{9a{\left(b+c\right)}^{2}}{64s\left(s-b\right)\left(s-c\right)}.\end{array}$
(2.11)

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

${\left({m}_{b}-{m}_{c}\right)}^{2}=\frac{{\left({m}_{b}^{2}-{m}_{c}^{2}\right)}^{2}}{{\left({m}_{b}+{m}_{c}\right)}^{2}}=\frac{9{\left(b+c\right)}^{2}{\left(b-c\right)}^{2}}{16{\left({m}_{b}+{m}_{c}\right)}^{2}}\le \frac{9}{4}{\left(b-c\right)}^{2}.$
(2.12)

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

${m}_{a}\ge \left\{\begin{array}{c}{m}_{1}\hfill \\ {m}_{b}\hfill \end{array}\ge {m}_{2}\ge {m}_{c}.$
(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}\left(\frac{{m}_{1}}{{m}_{2}+{m}_{b}}+\frac{{m}_{1}}{{m}_{2}+{m}_{c}}\right)-\frac{{m}_{2}}{{m}_{1}}-\frac{{m}_{1}}{4{m}_{2}}\\ \phantom{\rule{1em}{0ex}}=\frac{{m}_{b}+{m}_{c}-2{m}_{2}}{{m}_{a}+{m}_{1}}+\frac{{m}_{2}\left({m}_{1}-{m}_{a}\right)}{{m}_{1}\left({m}_{a}+{m}_{1}\right)}+\frac{{m}_{1}\left({m}_{2}^{2}-{m}_{b}{m}_{c}\right)}{4{m}_{2}\left({m}_{2}+{m}_{b}\right)\left({m}_{2}+{m}_{c}\right)}\\ \phantom{\rule{1em}{0ex}}\ge \frac{{\left({m}_{b}+{m}_{c}\right)}^{2}-4{m}_{2}^{2}}{\left({m}_{a}+{m}_{1}\right)\left({m}_{b}+{m}_{c}+2{m}_{2}\right)}+\frac{{m}_{2}\left({m}_{1}^{2}-{m}_{a}^{2}\right)}{{m}_{1}{\left({m}_{a}+{m}_{1}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}=\frac{2\left({m}_{b}^{2}+{m}_{c}^{2}\right)-{\left({m}_{b}-{m}_{c}\right)}^{2}-4{m}_{2}^{2}}{\left({m}_{a}+{m}_{1}\right)\left({m}_{b}+{m}_{c}+2{m}_{2}\right)}+\frac{{m}_{2}\left({m}_{1}^{2}-{m}_{a}^{2}\right)}{{m}_{1}{\left({m}_{a}+{m}_{1}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}=\frac{\frac{1}{4}{\left(b-c\right)}^{2}-{\left({m}_{b}-{m}_{c}\right)}^{2}}{\left({m}_{a}+{m}_{1}\right)\left({m}_{b}+{m}_{c}+2{m}_{2}\right)}-\frac{{m}_{2}{\left(b-c\right)}^{2}}{4{m}_{1}{\left({m}_{a}+{m}_{1}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}\ge \frac{\frac{1}{4}{\left(b-c\right)}^{2}-\frac{9}{4}{\left(b-c\right)}^{2}}{\left({m}_{a}+{m}_{1}\right)\left({m}_{b}+{m}_{c}+2{m}_{2}\right)}-\frac{{\left(b-c\right)}^{2}}{4{\left({m}_{a}+{m}_{1}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}=\frac{-2{\left(b-c\right)}^{2}}{\left({m}_{a}+{m}_{1}\right)\left({m}_{b}+{m}_{c}+2{m}_{2}\right)}-\frac{{\left(b-c\right)}^{2}}{4{\left({m}_{a}+{m}_{1}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}\ge \frac{-2{\left(b-c\right)}^{2}}{\left({m}_{a}+{m}_{1}\right)\left({m}_{b}+{m}_{c}\right)}-\frac{{\left(b-c\right)}^{2}}{4{\left({m}_{a}+{m}_{1}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}=\frac{-2{\left(b-c\right)}^{2}}{{\left({m}_{b}+{m}_{c}\right)}^{2}}-\frac{{\left(b-c\right)}^{2}}{4{\left({m}_{b}+{m}_{c}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}\ge \frac{-9{\left(b-c\right)}^{2}}{4{\left({m}_{b}+{m}_{c}\right)}^{2}}.\end{array}$
(2.14)

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

$\begin{array}{l}\frac{9a{\left(b+c\right)}^{2}}{64s\left(s-b\right)\left(s-c\right)}-\frac{9{\left(b+c\right)}^{2}}{8{\left({m}_{b}+{m}_{c}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}=\frac{9{\left(b+c\right)}^{2}\left[a{\left({m}_{b}+{m}_{c}\right)}^{2}-8s\left(s-b\right)\left(s-c\right)\right]}{64s\left(s-b\right)\left(s-c\right){\left({m}_{b}+{m}_{c}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}\ge \frac{9{\left(b+c\right)}^{2}}{64s\left(s-b\right)\left(s-c\right){\left({m}_{b}+{m}_{c}\right)}^{2}}\cdot \frac{3s\sqrt{\left(s-b\right)\left(s-c\right)}{\left(b-c\right)}^{2}}{a}\\ \phantom{\rule{1em}{0ex}}=\frac{27{\left(b+c\right)}^{2}{\left(b-c\right)}^{2}}{64a\sqrt{\left(s-b\right)\left(s-c\right)}{\left({m}_{b}+{m}_{c}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}\ge \frac{27{\left(b+c\right)}^{2}{\left(b-c\right)}^{2}}{32{a}^{2}{\left({m}_{b}+{m}_{c}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}\ge \frac{27{\left(b-c\right)}^{2}}{8{\left({m}_{b}+{m}_{c}\right)}^{2}}.\end{array}$
(2.15)

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

$\begin{array}{l}\left[\frac{{m}_{b}+{m}_{c}}{{m}_{a}+{m}_{1}}+\frac{1}{4}\left(\frac{{m}_{1}}{{m}_{2}+{m}_{b}}+\frac{{m}_{1}}{{m}_{2}+{m}_{c}}\right)-\frac{9{\left(b+c\right)}^{2}}{8{\left({m}_{b}+{m}_{c}\right)}^{2}}\right]\\ \phantom{\rule{2em}{0ex}}-\left[\frac{{m}_{2}}{{m}_{1}}+\frac{{m}_{1}}{4{m}_{2}}-\frac{9a{\left(b+c\right)}^{2}}{64s\left(s-b\right)\left(s-c\right)}\right]\\ \phantom{\rule{1em}{0ex}}=\left[\frac{{m}_{b}+{m}_{c}}{{m}_{a}+{m}_{1}}+\frac{1}{4}\left(\frac{{m}_{1}}{{m}_{2}+{m}_{b}}+\frac{{m}_{1}}{{m}_{2}+{m}_{c}}\right)-\frac{{m}_{2}}{{m}_{1}}-\frac{{m}_{1}}{4{m}_{2}}\right]\\ \phantom{\rule{2em}{0ex}}+\left[\frac{9a{\left(b+c\right)}^{2}}{64s\left(s-b\right)\left(s-c\right)}-\frac{9{\left(b+c\right)}^{2}}{8{\left({m}_{b}+{m}_{c}\right)}^{2}}\right]\\ \phantom{\rule{1em}{0ex}}\ge \frac{-9{\left(b-c\right)}^{2}}{4{\left({m}_{b}+{m}_{c}\right)}^{2}}+\frac{27{\left(b-c\right)}^{2}}{8{\left({m}_{b}+{m}_{c}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}=\frac{9{\left(b-c\right)}^{2}}{8{\left({m}_{b}+{m}_{c}\right)}^{2}}\ge 0.\end{array}$
(2.16)

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

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

$\frac{{m}_{2}}{{m}_{1}}+\frac{{m}_{1}}{4{m}_{2}}+\frac{3{\left(b+c\right)}^{2}}{16{a}^{2}}\ge 2$
(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.

1. (i)

First, we prove inequality (2.17).

$\begin{array}{rl}\frac{{m}_{2}}{{m}_{1}}+\frac{{m}_{1}}{4{m}_{2}}+\frac{3{\left(b+c\right)}^{2}}{16{a}^{2}}-2& =\sqrt{\frac{1+2{x}^{2}}{4-{x}^{2}}}+\frac{1}{4}\sqrt{\frac{4-{x}^{2}}{1+2{x}^{2}}}+\frac{3}{4{x}^{2}}-2\\ =\frac{8+7{x}^{2}}{4\sqrt{\left(4-{x}^{2}\right)\left(1+2{x}^{2}\right)}}+\frac{3\left(1-{x}^{2}\right)}{4{x}^{2}}-\frac{5}{4}\\ \ge \frac{8+7{x}^{2}}{4\cdot \frac{\left(4-{x}^{2}\right)+\left(1+2{x}^{2}\right)}{2}}+\frac{3\left(1-{x}^{2}\right)}{4{x}^{2}}-\frac{5}{4}\\ =\frac{8+7{x}^{2}}{2\left(5+{x}^{2}\right)}+\frac{3\left(1-{x}^{2}\right)}{4{x}^{2}}-\frac{5}{4}\\ =\frac{9\left({x}^{2}-1\right)}{4\left(5+{x}^{2}\right)}+\frac{3\left(1-{x}^{2}\right)}{4{x}^{2}}\\ \ge \frac{3\left({x}^{2}-1\right)}{8}+\frac{3\left(1-{x}^{2}\right)}{4}\\ =\frac{3\left(1-{x}^{2}\right)}{8}\ge 0.\end{array}$
(2.19)

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

1. (ii)

Second, we prove inequality (2.18).

$\begin{array}{l}{m}_{1}+\sqrt{3}a-\sqrt{3}s\\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}\sqrt{4-{x}^{2}}-\frac{\sqrt{3}}{2}\left(2-x\right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}\sqrt{2-x}\left(\sqrt{2+x}-\sqrt{3\left(2-x\right)}\right)\\ \phantom{\rule{1em}{0ex}}=\frac{-2\sqrt{2-x}\left(1-x\right)}{\sqrt{2+x}+\sqrt{3\left(2-x\right)}}\le 0.\end{array}$
(2.20)

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

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

${m}_{a}{m}_{b}+{m}_{b}{m}_{c}+{m}_{c}{m}_{a}-2{m}_{1}{m}_{2}-{m}_{2}^{2}\ge \frac{3}{8}{\left(b-c\right)}^{2}-\frac{3s\left(s-a\right){\left(b-c\right)}^{2}}{16\left(s-b\right)\left(s-c\right)}.$
(2.21)

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

${\left({m}_{b}+{m}_{c}\right)}^{2}\ge 4{m}_{b}{m}_{c}\ge 4s\sqrt{\left(s-b\right)\left(s-c\right)}\ge 6a\sqrt{\left(s-b\right)\left(s-c\right)}\ge 12\left(s-b\right)\left(s-c\right)$

or

${m}_{b}+{m}_{c}\ge 2\sqrt{3}\sqrt{\left(s-b\right)\left(s-c\right)}.$
(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}-2{m}_{1}{m}_{2}-{m}_{2}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{\left({m}_{b}+{m}_{c}\right)\left({m}_{a}^{2}-{m}_{1}^{2}\right)}{{m}_{a}+{m}_{1}}+\frac{{m}_{1}\left({m}_{b}^{2}-{m}_{2}^{2}\right)}{{m}_{b}+{m}_{2}}+\frac{{m}_{1}\left({m}_{c}^{2}-{m}_{2}^{2}\right)}{{m}_{c}+{m}_{2}}-\frac{{\left({m}_{b}^{2}-{m}_{c}^{2}\right)}^{2}}{2{\left({m}_{b}+{m}_{c}\right)}^{2}}+\frac{1}{16}{\left(b-c\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{\left({m}_{b}+{m}_{c}\right){\left(b-c\right)}^{2}}{4\left({m}_{a}+{m}_{1}\right)}+\frac{{m}_{1}\left(5b+7c\right)\left(c-b\right)}{16\left({m}_{b}+{m}_{2}\right)}+\frac{{m}_{1}\left(7b+5c\right)\left(b-c\right)}{16\left({m}_{c}+{m}_{2}\right)}\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{9{\left(b+c\right)}^{2}{\left(b-c\right)}^{2}}{32{\left({m}_{b}+{m}_{c}\right)}^{2}}+\frac{1}{16}{\left(b-c\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{\left({m}_{b}+{m}_{c}\right){\left(b-c\right)}^{2}}{4\left({m}_{a}+{m}_{1}\right)}+\frac{{m}_{1}{\left(b-c\right)}^{2}}{16\left({m}_{b}+{m}_{2}\right)}+\frac{{m}_{1}{\left(b-c\right)}^{2}}{16\left({m}_{c}+{m}_{2}\right)}\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{9{m}_{1}{\left(b+c\right)}^{2}{\left(b-c\right)}^{2}}{32\left({m}_{b}+{m}_{2}\right)\left({m}_{c}+{m}_{2}\right)\left({m}_{b}+{m}_{c}\right)}\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{9{\left(b+c\right)}^{2}{\left(b-c\right)}^{2}}{32{\left({m}_{b}+{m}_{c}\right)}^{2}}+\frac{1}{16}{\left(b-c\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{1}{4}\left(\frac{{m}_{2}}{{m}_{1}}+\frac{{m}_{1}}{4{m}_{2}}-\frac{9a{\left(b+c\right)}^{2}}{64s\left(s-b\right)\left(s-c\right)}-\frac{{m}_{1}{\left(b+c\right)}^{2}}{64\sqrt{3}s\left(s-b\right)\left(s-c\right)}+\frac{1}{4}\right){\left(b-c\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}\left(\frac{{m}_{2}}{{m}_{1}}+\frac{{m}_{1}}{4{m}_{2}}-\frac{9\left({m}_{1}+\sqrt{3}a\right){\left(b+c\right)}^{2}}{64\sqrt{3}s\left(s-b\right)\left(s-c\right)}+\frac{1}{4}\right){\left(b-c\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{1}{4}\left(\frac{{m}_{2}}{{m}_{1}}+\frac{{m}_{1}}{4{m}_{2}}-\frac{9{\left(b+c\right)}^{2}}{64\left(s-b\right)\left(s-c\right)}+\frac{1}{4}\right){\left(b-c\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}\left(\frac{{m}_{2}}{{m}_{1}}+\frac{{m}_{1}}{4{m}_{2}}-\frac{3\left[{\left(b+c\right)}^{2}-{a}^{2}\right]}{16\left(s-b\right)\left(s-c\right)}+\frac{3\left[{\left(b+c\right)}^{2}-4{a}^{2}\right]}{64\left(s-b\right)\left(s-c\right)}+\frac{1}{4}\right){\left(b-c\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{1}{4}\left(\frac{{m}_{2}}{{m}_{1}}+\frac{{m}_{1}}{4{m}_{2}}-\frac{3s\left(s-a\right)}{4\left(s-b\right)\left(s-c\right)}+\frac{3\left[{\left(b+c\right)}^{2}-4{a}^{2}\right]}{16{a}^{2}}+\frac{1}{4}\right){\left(b-c\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}\left(\frac{{m}_{2}}{{m}_{1}}+\frac{{m}_{1}}{4{m}_{2}}+\frac{3{\left(b+c\right)}^{2}}{16{a}^{2}}-\frac{3s\left(s-a\right)}{4\left(s-b\right)\left(s-c\right)}-\frac{1}{2}\right){\left(b-c\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{1}{4}\left(2-\frac{3s\left(s-a\right)}{4\left(s-b\right)\left(s-c\right)}-\frac{1}{2}\right){\left(b-c\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{3}{8}{\left(b-c\right)}^{2}-\frac{3s\left(s-a\right){\left(b-c\right)}^{2}}{16\left(s-b\right)\left(s-c\right)}.\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 $x\in \left(0,1\right]$, then inequality (1.1) is equivalent to

$\begin{array}{r}2{\left(\frac{x\sqrt{4-{x}^{2}}}{2\left(2-x\right)}-\frac{\sqrt{4-{x}^{2}}}{2}\right)}^{2}\ge 2k{\left(\frac{\sqrt{4-{x}^{2}}}{2}-\frac{\sqrt{2{x}^{2}+1}}{2}\right)}^{2}\\ \phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}\frac{2+x}{2-x}\ge k\cdot \frac{9{\left(1+x\right)}^{2}}{4{\left(\sqrt{4-{x}^{2}}+\sqrt{2{x}^{2}+1}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}k\le \frac{4\left(2+x\right){\left(\sqrt{4-{x}^{2}}+\sqrt{2{x}^{2}+1}\right)}^{2}}{9\left(2-x\right){\left(1+x\right)}^{2}}.\end{array}$
(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, then we have

$\sum {\left({r}_{b}-{r}_{c}\right)}^{2}-k\cdot \sum {\left({m}_{b}-{m}_{c}\right)}^{2}\ge 2{\left({r}_{1}-{m}_{1}\right)}^{2}-2k{\left({m}_{1}-{m}_{2}\right)}^{2}.$
(2.24)

Proof For

$\sum {\left({r}_{b}-{r}_{c}\right)}^{2}=2\sum {r}_{a}^{2}-2\sum {r}_{b}{r}_{c}=2\sum {r}_{a}^{2}-2{s}^{2}$

and

$\sum {\left({m}_{b}-{m}_{c}\right)}^{2}=2\sum {m}_{a}^{2}-2\sum {m}_{b}{m}_{c}=\frac{3}{2}\sum {a}^{2}-2\sum {m}_{b}{m}_{c},$

hence, by Lemmas 2.1 and 2.6, we have

$\begin{array}{c}\sum {\left({r}_{b}-{r}_{c}\right)}^{2}-k\cdot \sum {\left({m}_{b}-{m}_{c}\right)}^{2}-2{\left({r}_{1}-{m}_{1}\right)}^{2}+2k{\left({m}_{1}-{m}_{2}\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=2\left[\sum {r}_{a}^{2}-{r}_{1}^{2}-2{m}_{1}^{2}\right]+2k\left[\sum {m}_{b}{m}_{c}-2{m}_{1}{m}_{2}-{m}_{2}^{2}-\frac{3}{8}{\left(b-c\right)}^{2}\right]\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{3s\left(s-a\right){\left(b-c\right)}^{2}}{2\left(s-b\right)\left(s-c\right)}-\frac{3ks\left(s-a\right){\left(b-c\right)}^{2}}{8\left(s-b\right)\left(s-c\right)}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{3\left(4-k\right)s\left(s-a\right){\left(b-c\right)}^{2}}{8\left(s-b\right)\left(s-c\right)}\ge 0.\hfill \end{array}$

The proof of Lemma 2.8 is complete. □

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

Define

$F\left(x\right)={a}_{0}{x}^{n}+{a}_{1}{x}^{n-1}+\cdots +{a}_{n},$

and

$G\left(x\right)={b}_{0}{x}^{m}+{b}_{1}{x}^{m-1}+\cdots +{b}_{m}.$

If ${a}_{0}\ne 0$ or ${b}_{0}\ne 0$, then the polynomials $F\left(x\right)$ and $G\left(x\right)$ have a common root if and only if

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

Lemma 2.10 (see [7, 8])

Given a polynomial $f\left(x\right)$ with real coefficients

$f\left(x\right)={a}_{0}{x}^{n}+{a}_{1}{x}^{n-1}+\cdots +{a}_{n},$

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

$\left\{{D}_{1}\left(f\right),{D}_{2}\left(f\right),\dots ,{D}_{n}\left(f\right)\right\}$

is v, then the number of the pairs of distinct conjugate imaginary roots of $f\left(x\right)$ 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 $f\left(x\right)$ 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.

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

$2{\left({r}_{1}-{m}_{1}\right)}^{2}-2k{\left({m}_{1}-{m}_{2}\right)}^{2}\ge 0$

or, equivalently, that

$\begin{array}{l}{\left(\frac{a\sqrt{{\left(b+c\right)}^{2}-{a}^{2}}}{2\left(b+c-a\right)}-\frac{\sqrt{{\left(b+c\right)}^{2}-{a}^{2}}}{2}\right)}^{2}-k{\left(\frac{\sqrt{{\left(b+c\right)}^{2}-{a}^{2}}}{2}-\frac{1}{2}\sqrt{2{a}^{2}+\frac{1}{4}{\left(b+c\right)}^{2}}\right)}^{2}\\ \phantom{\rule{1em}{0ex}}\ge 0.\end{array}$
(3.1)

Without loss of generality, we can assume that

$a=t\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{b+c}{2}=1\phantom{\rule{1em}{0ex}}\left(0

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:

${\left(\frac{t\sqrt{4-{t}^{2}}}{2\left(2-t\right)}-\frac{\sqrt{4-{t}^{2}}}{2}\right)}^{2}-k{\left(\frac{\sqrt{4-{t}^{2}}}{2}-\frac{\sqrt{2{t}^{2}+1}}{2}\right)}^{2}\ge 0.$
(3.2)

We consider the following two cases separately.

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

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

$k\le \frac{4\left(2+t\right){\left(\sqrt{4-{t}^{2}}+\sqrt{2{t}^{2}+1}\right)}^{2}}{9\left(2-t\right){\left(1+t\right)}^{2}}.$
(3.3)

Define the function

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

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

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

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

$\sqrt{4-{t}^{2}}\cdot \left[\left(2{t}^{3}+5{t}^{2}+10t-2\right)-\left(2-t\right)\sqrt{4-{t}^{2}}\cdot \sqrt{2{t}^{2}+1}\right]=0.$
(3.4)

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

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

that is,

${\left(1+t\right)}^{2}\phi \left(t\right)=0,$
(3.5)

where

$\phi \left(t\right)={t}^{4}+10{t}^{2}-2.$

It is obvious that the equation

${\left(1+t\right)}^{2}=0$
(3.6)

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

It is easy to find that the equation

$\phi \left(t\right)=0$
(3.7)

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

$g{\left(t\right)}_{\mathrm{max}}=g\left({t}_{0}\right)\approx 3.2817755127\in \left(3,4\right).$
(3.8)

Now we prove $g\left({t}_{0}\right)$ is the root of equation (1.2). For this purpose, we consider the following nonlinear algebraic equation system:

$\left\{\begin{array}{c}\phi \left({t}_{0}\right)=0,\hfill \\ 2{t}_{0}^{2}+1-{u}_{0}^{2}=0,\hfill \\ 4-{t}_{0}^{2}-{v}_{0}^{2}=0,\hfill \\ 4\left(2+t\right){\left({u}_{0}+{v}_{0}\right)}^{2}-9\left(2-t\right){\left(1+t\right)}^{2}k=0.\hfill \end{array}$
(3.9)

It is easy to see that $g\left({t}_{0}\right)$ 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

$29\text{,}648\text{,}323\text{,}021\text{,}629\text{,}456\cdot {\varphi }_{1}^{2}\left(k\right)\cdot {\varphi }_{2}^{2}\left(k\right)=0,$
(3.10)

where

${\varphi }_{1}\left(k\right)=6\text{,}561{k}^{4}-14\text{,}256{k}^{3}-18\text{,}080{k}^{2}-25\text{,}344k+20\text{,}736$

and

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

The revised sign list of the discriminant sequence of ${\varphi }_{1}\left(k\right)$ is given by

$\left[1,1,-1,-1\right].$
(3.11)

The revised sign list of the discriminant sequence of ${\varphi }_{2}\left(k\right)$ is given by

$\left[1,1,-1,-1\right].$
(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

${\varphi }_{1}\left(k\right)=0$
(3.13)

and

${\varphi }_{2}\left(k\right)=0$
(3.14)

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

$\begin{array}{c}{\varphi }_{1}\left(0\right)=20\text{,}736>0;\phantom{\rule{2em}{0ex}}{\varphi }_{2}\left(0\right)=2\text{,}304>0,\hfill \\ {\varphi }_{1}\left(1\right)=-30\text{,}383<0;\phantom{\rule{2em}{0ex}}{\varphi }_{2}\left(1\right)=-8\text{,}567<0,\hfill \\ {\varphi }_{1}\left(3\right)=-71\text{,}487<0;\phantom{\rule{2em}{0ex}}{\varphi }_{2}\left(8\right)=-313\text{,}088<0\hfill \end{array}$

and

${\varphi }_{1}\left(4\right)=397\text{,}312>0;\phantom{\rule{2em}{0ex}}{\varphi }_{2}\left(9\right)=26\text{,}793>0.$

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

$\left(0,1\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(3,4\right).$

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

$\left(0,1\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(8,9\right).$

Hence, by (3.8), we can conclude that $g\left({t}_{0}\right)$ is the root of equation (1.2). The proof of Theorem 1.1 is thus completed. □

## References

1. Liu B-Q: BOTTEMA, What We Have Seen - the New Theory, New Method and New Result of the Research on Triangle. Tibet People Press, Lhasa; 2003. (in Chinese)

2. Yang L: A dimension-decreasing algorithm with generic program for automated inequality proving. High Technol. Lett. 1998, 25(7):20–25. (in Chinese)

3. Yang L: Recent advances in automated theorem proving on inequalities. J. Comput. Sci. Technol. 1999, 14(5):434–446. 10.1007/BF02948785

4. Yang L, Xia B-C: Automated Proving and Discovering on Inequalities. Science Press, Beijing; 2008. (in Chinese)

5. Yang L, Xia S-H: Automated proving for a class of constructive geometric inequalities. Chinese J. Comput. 2003, 26(7):769–778. (in Chinese)

6. Sylvester JJ: A method of determining by mere inspection the derivatives from two equations of any degree. Philos. Mag. 1840, 16: 132–135.

7. Yang L, Zhang J-Z, Hou X-R: Nonlinear Algebraic Equation System and Automated Theorem Proving. Shanghai Scientific and Technological Education Press, Shanghai; 1996:23–25. (in Chinese)

8. Yang L, Hou X-R, Zeng Z-B: A complete discrimination system for polynomials. Sci. China Ser. E 1996, 39(7):628–646.

## Acknowledgements

The authors would like to thank the anonymous referees for their very careful reading and making some valuable comments which have essentially improved the presentation of this paper.

## Author information

Authors

### Corresponding author

Correspondence to Shi-Chang Shi.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

Both authors contributed equally and read and approved the final manuscript.

## Rights and permissions

Reprints and permissions

Shi, SC., Wu, YD. An artificial proof of a geometric inequality in a triangle. J Inequal Appl 2013, 329 (2013). https://doi.org/10.1186/1029-242X-2013-329