An inequality with a prime number I have this to purpose :

Let $a,b,c$ be real positive numbers such that $abc=1$ prove that :
  $$\frac{a^2}{(a^{11}+1)^2}+\frac{b^2}{(b^{11}+1)^2}+\frac{c^2}{(c^{11}+1)^2}\leq \frac{10^7}{9193531}$$

Where $9193531$ is a prime number wich makes the problem harder . 
Really I have no ideas to prove this and all my classical methods fails automatically when I want solve this problem . 
Furthermore I do not know which class of functions belongs $f(x)$ where :
$$f(x)=\frac{x^2}{(x^{11}+1)^2}$$
Maybe we can find a way with the concept of Quasiconvex function 
If you have any hints it would be nice.
Thanks 

Geometric Explanation "min/max at x=y"
(Not an Answer) 
  
 
Using $c=1/(ab)$, Let $a=x$, $b=y$ and define:
$$ 
\begin{align} 
&f(x,y)=\frac{10^7}{9193531}-\frac{x^{2}}{\left(x^{11}+1\right)^2}-\frac{y^{2}}{\left(y^{11}+1\right)^2}-\frac{(xy)^{20}}{\left((xy)^{11}+1\right)^2}=f(y,x) \\ 
&f({\small-\frac{8}{10}},{\small-\frac{9}{10}})\approx-1.4\lt0, \quad f({\small2},{\small2})\approx+1.0\gt0 
\end{align} 
$$ 
 
Benefiting from the function symmetric behavior $f(x,y)=f(y,x)$, the inequality holds for the special case min/max accord at $x=y$. Thus, re-define: 
$$ 
\begin{align} 
&f(x)=\frac{10^7}{9193531}-\frac{2x^{2}}{\left(x^{11}+1\right)^2}-\frac{x^{40}}{\left(x^{22}+1\right)^2} \\ 
&\qquad=\color{blue}{\frac{10^7}{9193531}-\frac{2x^{2}+4x^{24}+x^{40}+2x^{46}+2x^{51}+x^{62}}{1+2x^{11}+3x^{22}+4x^{33}+3x^{44}+2x^{55}+x^{66}}} 
\end{align} 
$$ 
And the question now turns to be about $\,{\small f(x)}\,$ roots for $\,{\small x\gt0}\,$.



 
Putting $\,{\small f(x)=0}\,$ and simplifying, gives the equivalent polynomial: 
$$ 
\begin{align} 
p(x)=\,&\,\color{red}{{10^7}\left({1+2x^{11}+3x^{22}+4x^{33}+3x^{44}+2x^{55}+x^{66}}\right)} \\ 
&\,\color{red}{-{9193531}\left(2x^{2}+4x^{24}+x^{40}+2x^{46}+2x^{51}+x^{62}\right)} 
\end{align} 
$$ 
Because the polynomial degree is high, and to determine if $\,{\small p(x)}\,$ has any $\,{\color{red}{\small\text{real root}}}\,$ for $\,{\small x\gt0}\,$, it is enough to use Sturm theorem to compare number of roots inside the two regions $\,{\small [-1,0]}\,$ and $\,{\small [-1,\infty)}\,$, which should result in only $\,{\color{red}{\small\text{ONE}}}\,$ root in both, hence $\,{\small p(x)\ne0\,\,\colon\,x\in(0,\infty)}\,$. 
  

