Prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}$ for $a, b, c > 0$ with $abc = 1$ 
Let $a$, $b$ and $c$ be positive numbers such  that $abc=1$. Prove that:
$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq\frac{3}{2}.$$

I tried homogenization and the BW (https://artofproblemsolving.com/community/c6h522084),
but it does not work.
Indeed, let $a=\frac{x}{y}$, $b=\frac{y}{z}$, where $x$, $y$ and $z$ are positives.
Hence, $c=\frac{z}{x}$ and we need to prove that $$\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}\leq\frac{3}{2},$$
which has a problem around $(x,y,z)=(7,5,6)$.
For these values $$\frac{3}{2}-\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}=0.0075...$$
I tried also TL, uvw, C-S, Lagrange multipliers and more, but without success.
Also, Vasc's Theorems don't help.
Also, the following method does not help here. Find the maximum of the expression
Because the inequality $\frac{x}{x^{11}+1}\leq\frac{3(a^9+1)}{4(a^{18}+a^9+1)}$ is wrong.

In principle it is, if not always, almost very often that such a problem can be solved using techniques from optimization. For instance one can consider the following maximization problem:
\begin{equation}
\max_{a,b,c\in C} f(a,b,c) 
\end{equation}
where the constraint set $C:=\{a,b,c\in\mathbb{R}_+:abc=1\}$ and $f(a,b,c):=\sum_{cyc}a/(a^{11}+1)$. If one shows that $3/2$ is the maximum value $f(a,b,c)$ attains in $C$ then this solves the inequality problem. We will follow the same idea however first we transform the given inequality into an equivalent form together with an appropriate constraint which makes it easier to solve it as a maximization problem. The original inequality is given by:
\begin{equation}
\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leqslant \frac{3}{2}
\end{equation}
and $abc=1$. After proper rearrangements this last inequality is equivalent to:
$$2(a(b^{11}+1)(c^{11}+1)+b(a^{11}+1)(c^{11}+1)+c(a^{11}+1)(b^{11}+1))\leqslant 3(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
or equivalently:
$$2\Big((b^{10}+\frac{1}{b})(c^{10}+\frac{1}{c})+(a^{10}+\frac{1}{a})(c^{10}+\frac{1}{c})+(a^{10}+\frac{1}{a})(b^{10}+\frac{1}{b})\Big)\leqslant 3(a^{10}+\frac{1}{a})(b^{10}+\frac{1}{b})(c^{10}+\frac{1}{c})$$
Let $f(x):=x^{10}+1/x$ then the last inequality is the same as:
$$\frac{1}{f(a)}+\frac{1}{f(b)}+\frac{1}{f(c)}\leqslant\frac{3}{2}$$
It is sufficient to look at the problem:
$$\max_{a,b,c}F(a,b,c):=\frac{1}{f(a)}+\frac{1}{f(b)}+\frac{1}{f(c)}$$
subject to $abc=1$. The Lagrangian for this problem is: 
$$L(a,b,c,\lambda):=F(a,b,c)-\lambda(1-abc)$$
From the first order conditions we get the following equations:
$$\frac{f'(a)}{f^2(a)}=\lambda bc\Leftrightarrow a\frac{f'(a)}{f^2(a)}=\lambda \\
\frac{f'(b)}{f^2(b)}=\lambda ac\Leftrightarrow b\frac{f'(b)}{f^2(b)}=\lambda \\
\frac{f'(c)}{f^2(c)}=\lambda ab\Leftrightarrow c\frac{f'(c)}{f^2(c)}=\lambda $$
A possible obvious solution to this system is $a=b=c=1$ and $\lambda=9/4$. If one calculates the Hessian of $L(a,b,c,\lambda)$ (bordered Hessian) we get:
$$\text{Hess}_L(a,b,c,\lambda)=
 \begin{bmatrix}
 0 & bc & ac & ab\\
bc & F_{aa} & \lambda c& \lambda b \\
ac &\lambda c & F_{bb} & \lambda a\\
ab &\lambda b & \lambda a & F_{cc} 
\end{bmatrix}
$$
where $F_{xx}:=-\Big[(f''(x)f^2(x)-2(f'(x))^2f(x))/f^4(x)\Big]$. Evaluating at $(1,1,1,9/4)$ gives:
$$\text{Hess}_L(1,1,1,9/4)=
 \begin{bmatrix}
 0 & 1 & 1 & 1\\
1 & -11/4 & 9/4& 9/4 \\
1 &9/4 & -11/4 & 9/4\\
1 &9/4 & 9/4 & -11/4 
\end{bmatrix}
$$
From this follows $\det \text{Hess}_L(1,1,1,9/4)=-75$ and $\det M_L(1,1,1,9/4)=10$ where $M$ is the submatrix
$$M_L:=\begin{bmatrix}
 0 & 1 & 1 \\
1 & -11/4 & 9/4 \\
1 &9/4 & -11/4
\end{bmatrix}
$$
These satisfy the second order conditions for our bordered Hessian (the alternating sign condition) for a local maximum at $(1,1,1)$. 
