Inequality $\sum_{cyc}\frac{a^{60}}{\frac{11}{2}a^{50}+5b^{50}}\geq 2\frac{a^{10}+b^{10}+c^{10}}{21}$ Inspired by some questions on Math.stack I propose this :


Let $a,b,c>0$ then we have :
    $$\sum_{cyc}\frac{a^{60}}{\frac{11}{2}a^{50}+5b^{50}}\geq 2\frac{a^{10}+b^{10}+c^{10}}{21}$$ 


First I try Am-Gm but it's a wrong way .I try also Tchebytchev's inequality but it's not sufficient . Furthermore we cannot invoke convexity .I'm not sure that we can use Buffalo's way as here.Finally I can prove that :
$$\sum_{cyc}\frac{a^{60}}{\frac{11}{2}a^{50}+5b^{50}}+\sum_{cyc}\frac{b^{60}}{\frac{11}{2}b^{50}+a^{50}}\geq 4\frac{a^{10}+b^{10}+c^{10}}{21}$$
But I can't prove (adding an order on $a,b,c$) that we have :
$$\sum_{cyc}\frac{a^{60}}{\frac{11}{2}a^{50}+5b^{50}}-\sum_{cyc}\frac{b^{60}}{\frac{11}{2}b^{50}+a^{50}}\geq 0$$
Maybe my new method could work I have not try .
My question :
How to prove it ?
Thanks in advance to your help.
 A: We need to prove that:
$$\sum_{cyc}\frac{a^6}{11a^5+10b^5}\geq\frac{a+b+c}{21},$$ where $a$, $b$ and $c$ are positives.
Indeed, since our inequality is cyclic, we can assume that $a=\min\{a,b,c\}$.
Now, let $b=a+u$ and $c=a+v$.
Thus, $u$ and $v$ are non-negatives and we obtain something smooth: 
https://www.wolframalpha.com/input/?i=%28x%5E6%2F%2811x%5E5%2B10y%5E5%29%2By%5E6%2F%2811y%5E5%2B10z%5E5%29%2Bz%5E6%2F%2811z%5E5%2B10x%5E5%29-%28x%2By%2Bz%29%2F21%2921%2811x%5E5%2B10y%5E5%29%2Cx%3Da%2Cy%3Da%2Bu%2Cz%3Da%2Bv
Only two polynomials maybe negative: the coefficient before $a^{10}$ and the coefficient before $a^{9}$, but it not so disturbs.
Can you end it now?
Finally, it's enough to prove that:
$$1680t^{14}+13316t^{13}+38025t^{12}+45029t^{11}-5161t^{10}-57574t^9+21367t^8+210530t^7+...\geq0$$ for $t>0$,  which is obvious.
The following inequality is also true.

Let $a$, $b$ and $c$ be positive numbers. Prove that:
  $$\frac{a^6}{9a^5+8b^5}+\frac{b^6}{9b^5+8c^5}+\frac{c^6}{9c^5+8a^5}\geq\frac{a+b+c}{17}.$$

$$\sum_{cyc}\frac{a^6}{8a^5+7b^5}\geq\frac{a+b+c}{15}$$ is wrong already.
