A nice Nesbitt inequality from a strange inequality 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$!$ 

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