A nice Nesbitt inequality from a strange inequality

Question

Given $a,\,b,\,c> 0$$,$ prove that$:$ $$\frac{a}{b+ c}+ \frac{b}{c+ a}+ \frac{c}{a+ b}+ \frac{63}{5}\left [ \frac{2\,c^{\,2}}{(\,a+ b\,)^{\,2}}- \frac{c}{a+ b} \right ]\geqq 0$$ See$:$ $\lceil$ https://artofproblemsolving.com/community/c6h354642p1923888 $\rfloor$

The only way I tried is Buffalo Way but the coefficient of$:$ $\text{coef}[\,c^{\,4}\,],\,\text{coef}[\,c^{\,3}\,],\,\text{coef}[\,c^{\,2}\,],\,\text{coef}[\,c\,],\,\text{coef}[\,c^{\,0}\,]$ aren$'$t same non$-$negative$,$ maybe I$'$m wrong because the equality occurs with $a= b$$.$ So I ask$,$ hope to see the best way$!$ Good luck everybody$!$

Answer 1

Let $\frac{c}{a+b}=\frac{x}{2}$.

Thus, by C-S and AM-GM $$\sum_{cyc}\frac{c}{a+b}+\frac{63}{5}\left(\frac{2c^2}{(a+b)^2}-\frac{c}{a+b}\right)=\frac{a}{b+c}+\frac{b}{a+c}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)=$$ $$=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)\geq\frac{(a+b)^2}{2ab+c(a+b)}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)=$$ $$=\frac{1}{\frac{2ab}{(a+b)^2}+\frac{c}{a+b}}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)\geq\frac{1}{\frac{1}{2}+\frac{x}{2}}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right).$$ Id est, it's enough to prove that $$\frac{1}{\frac{1}{2}+\frac{x}{2}}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)\geq0$$ or $$63x^3+5x^2-58x+20\geq0,$$ which is obvious.

Answer 2

We have: $$f\left(a,b,c\right)=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{63}{5}\left[\frac{2c^2}{(a+b\, )^2}-\frac{c}{a+b}\right]\ge 0$$

Let $t=\dfrac{a+b}2>0$. We hope that $f(a,b,c)\ge f(t;t;c)$. Indeed,

$$f(a,b,c)-f\left(t,t,c\right)=\frac{\left(a-b\right)^2\left(a+b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b+2c\right)}\ge 0$$

So we only prove $f(t,t,c)\ge0$

Or $$63c^3+5c^2t-58ct^2+20t^3\ge 0$$