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.

Define
$$
f(a,\lambda) = -\frac{a}{a^{11}+1} + \lambda \log(a) + \frac{1}{2}
$$
Then, for any choice of $\lambda$, 
$$
f(a,\lambda) + f(b,\lambda) + f(c,\lambda) = -\frac{a}{a^{11}+1} -\frac{b}{b^{11}+1} -\frac{c}{c^{11}+1} + \frac{3}{2}
$$
and we need to show that this is $\ge 0$.
It suffices to show that, for some $\lambda^*$ and for all $a$, $f(a, \lambda^*) \ge 0$.
Clearly, for any lambda, $f(a=1,\lambda)  = 0$. In order to keep $f(a,\lambda) $ positive for $a >1$ and $a <1$, we demand 
$$
0 = \frac{d f(a,\lambda)}{d a}|_{a=1}
$$
which results in $\lambda^* = - \frac94$. We therefore investigate  
$$
f(a,\lambda^*) = -\frac{a}{a^{11}+1} -\frac{9}{4} \log(a) + \frac{1}{2}
$$
By inspection, we have that  $f(a,\lambda^*) \ge 0$ for $a\in (0, 1.1]$. So the inequality is obeyed at least for $a,b,c < 1.1$, and it remains to be shown  that the inequality is obeyed outside this specification.
This gives rise to three cases:
case 1: $a,b,c > 1.1$. This is not possible since $abc = 1$.
case 2: $a < 1.1$ ; $b,c > 1.1$.  Now observe two facts:


*

*By inspection, $ \frac{a}{a^{11}+1} < 0.75$ for any $a$. 

*For  $b > 1.1$, $ \frac{b}{b^{11}+1} \le \frac{1.1}{1.1^{11}+1} \simeq 0.2855$ since $ \frac{b}{b^{11}+1}$ is falling for $b > 1.1$. 
Hence, in case 2, $ \frac{a}{a^{11}+1} + \frac{b}{b^{11}+1}+ \frac{c}{c^{11}+1} < 0.75 + 2\cdot 0.2855 = 1.3210 < \frac32$ which proves case 2.
case 3: $a,b < 1.1$ ; $c > 1.1$. Here $abc = 1$ requires $a\cdot b =1/c < 1.1^{-1} = 0.909$. Also note that, for some given $c$, $1/(1.1 c)  <a<1.1$ in order to observe $a,b < 1.1$.  Following case 2, we have that $f(c) = \frac{c}{c^{11}+1} $ is falling with $c$. These conditions could be further exploited (this has not yet been pursued in the comments).
As  Martin R. poined out, the maximum will be attained at a point where at least two out of $a,b,c$ equal. In this case, this would be $a=b$. So we can consider proving
$$
g(a) =  \frac32 - \frac{2 a}{a^{11}+1} - \frac{a^{-2}}{a^{-22}+1} \ge 0
$$
for $a < 1/\sqrt{1.1} \simeq 0.9535$. 
Note that in this range, the minimum of $g(a)$ occurs at $a^*\simeq 0.8385$ and has a value of $g(a^*) \simeq 0.00525$. Other than this inspection of the function $g(a)$, I couldn't offer a better proof.  
