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$!$ 

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. 
