Challenging inequality with three variables I'm interested by the following problem :

Let $a,b,c$ be real positive numbers such that $abc=1$ with and $\beta>1$ and $0<\alpha<1$  then  :
  $$\Big(\frac{\alpha a}{a^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha b}{b^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha c}{c^{11}+1}\Big)^\frac{1}{\beta}\leq 3\Big(\frac{\alpha}{2}\Big)^\frac{1}{\beta}$$

I claim that the maximum is reached for the triplet $(1;1;1)$ 
But I can't prove it ..
Any helps or hints would be appreciated .
Edit :
We start with the case $a\leq 1$ , $b\leq 1$ , $c\geq 1$ so we have :
$$\Big(\frac{\alpha a}{a^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha b}{b^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha c}{c^{11}+1}\Big)^\frac{1}{\beta}$$
Or with $a\geq 1$, $b\geq 1$ , $c\leq 1$ :
$$\Big(\frac{\alpha a^{10}}{a^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha b^{10}}{b^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha c^{10}}{c^{11}+1}\Big)^\frac{1}{\beta}$$
We have the following lemma :

Let $a,b$ be real positive numbers with $a\geq 1$, $b\geq 1$ then we have :
  $$\Big(\frac{\alpha a^{10}}{a^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha b^{10}}{b^{11}+1}\Big)^\frac{1}{\beta}\leq2\Big(\frac{\frac{\alpha a^{10}}{a^{11}+1}+\frac{\alpha b^{10}}{b^{11}+1}}{2}\Big)^\frac{1}{\beta}\leq 2\Big(\frac{a+b}{2ab}\frac{\alpha(\frac{2ab}{a+b})^{11}}{(\frac{2ab}{a+b})^{11}+1}\Big)^\frac{1}{\beta}$$

Proof :
It's just the inequality of Jensen apply to $f(x)$ wich is concave for $x\geq 1$ :
$f(x)=\Big(\frac{\alpha x^{11}}{x^{11}+1}\Big)$
The variable are :
$x_1=a$ and $x_2=b$
With coefficient :
$\alpha_1=\frac{1}{a}\frac{ab}{a+b}$
And 
$\alpha_2=\frac{1}{b}\frac{ab}{a+b}$
We have this other lemma :

$$\Big(\frac{\alpha c^{10}}{c^{11}+1}\Big)^\frac{1}{\beta}=\Big(\frac{\alpha ab}{(ab)^{11}+1}\Big)^\frac{1}{\beta}\leq \Big(\frac{\alpha (\frac{2ab}{a+b})^{2}}{(\frac{2ab}{a+b})^{22}+1}\Big)^\frac{1}{\beta} $$

Proof :
It's easy to show this because $f(x)=\Big(\frac{\alpha x}{x^{11}+1}\Big)$ is decreasing for $x\geq 1$
It's remains to prove :
$$(\frac{2ab}{a+b})^{2}\leq ab $$
Wich is obvious.
So we have  :
$$2\Big(\frac{a+b}{2ab}\frac{\alpha( \frac{2ab}{a+b})^{11}}{(\frac{2ab}{a+b})^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha( \frac{2ab}{a+b})^{2}}{(\frac{2ab}{a+b})^{22}+1}\Big)^\frac{1}{\beta}$$
Now we put :
$x=\frac{2ab}{a+b}$
We get for $x\geq 1$:
$$2\Big(\frac{\alpha (x)^{10}}{(x)^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha (x)^{2}}{(x)^{22}+1}\Big)^\frac{1}{\beta}\leq 3\Big(\frac{\alpha}{2}\Big)^\frac{1}{\beta}$$
My questions  : 
How to get the other cases ? 
How to prove this last one variable inequality ? 
Have you another way to prove this ?
 A: $\alpha$ can simply be factored out, so we can set it to $1$ and then ignore it.

At Least Two of $\boldsymbol{a,b,c}$ Must be Equal
To insure that
$$
\delta\left[f(a)+f(b)+f(c)\right]=f'(a)\,\delta a+f'(b)\,\delta b+f'(c)\,\delta c=0
$$
for all variations $\delta a,\delta b,\delta c$ so that
$$
\delta(abc)=abc\left(\frac{\delta a}{a}+\frac{\delta b}{b}+\frac{\delta c}{c}\right)=0
$$
orthogonality requires a $\lambda$ so that
$$
af'(a)=bf'(b)=cf'(c)=\lambda
$$
For any $n\in\mathbb{N}$ and $f(x)=\left(\frac{x}{x^{11}+1}\right)^{1/\beta}$, if we look at $xf'(x)$, we see that it is $2$-$1$ everywhere, except at the extreme points. This means that whatever $\lambda$ we have, there are at most two values of $x$ so that $xf'(x)=\lambda$. That is, at least two of $a,b,c$ must be equal.

Optimal Value of $\boldsymbol{a}$
Without loss of generality, assume that $b=a$ and $c=a^{-2}$. Then we want to maximize $2f(a)+f\!\left(a^{-2}\right)$:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}a}\left(2f(a)+f\!\left(a^{-2}\right)\right)
&=\frac2a\left(af'(a)-a^{-2}f'\!\left(a^{-2}\right)\right)\\
&=0
\end{align}
$$
is true when $a=1$. However, if $1\lt\beta\lt1.088$, there are two values of $a$ where the derivative vanishes. Looking at plots, it appears that the critical point at $a=1$ gives the maximum. In fact, if we look at the plot for $\beta=1$,  the critical point at $a=1$ gives the maximum:

Greater values of $\beta$ give a smaller maximum for $a\lt1$.
If we accept that $a=b=c=1$ gives the maximum, we get
$$
\left(\frac{a}{a^{11}+1}\right)^{1/\beta}+\left(\frac{b}{b^{11}+1}\right)^{1/\beta}+\left(\frac{c}{c^{11}+1}\right)^{1/\beta}\le3\left(\frac12\right)^{1/\beta}
$$
