$\frac{a}{a^a+1}+\frac{b}{b^b+1}+\frac{c}{c^c+1}\leq \frac{3}{2}$ with $abc=1$ 
Let $a,b,c>0$ such that $abc=1$ then we have :
  $$\frac{a}{a^a+1}+\frac{b}{b^b+1}+\frac{c}{c^c+1}\leq \frac{3}{2}$$

My try :
The original inequality is equivalent to :
$$a(b^b+1)(c^c+1)+b(a^a+1)(c^c+1)+c(a^a+1)(b^b+1)\leq \frac{3}{2}(a^a+1)(c^c+1)(b^b+1)$$
Or :
$$(a-\frac{3}{2}(a^a+1))(b^b+1)(c^c+1)+(b-\frac{3}{2}(b^b+1))(a^a+1)(c^c+1)+(c-\frac{3}{2}(c^c+1))(a^a+1)(b^b+1)\leq0$$
The function :
$$f(x)=x-\frac{3}{2}(x^{x}+1)$$
Is concave so we can apply Jensen's inequality but it's very ugly !
So have you an alternative proof ?
Thanks in advance for your time .
 A: Let $f(x)=\frac{x}{x^x+1}.$
We have $$f'(x)=\frac{1+(1-x-x\ln{x})x^x}{(x^x+1)^2},$$
which gives that $f'(x_1)=0$, where $x_1=1.322...$ and $f$ increases on$(0,x_1]$ and decreases on $[x_1,+\infty).$
Thus, $$\max_{(0,+\infty)}f=f(x_1)=0.54...$$
Also, $$\frac{3}{2}-\sum_{cyc}\frac{a}{a^a+1}=\sum_{cyc}\left(\frac{1}{2}-\sum_{cyc}\frac{a}{a^a+1}+\frac{1}{4}\ln{a}\right)\geq0$$
for all $\min\{a,b,c\}\geq x_0=0.23356...$
Let $a<x_0$.
Thus, $$\sum_{cyc}\frac{a}{a^a+1}<f(x_0)+2f(x_1)=1.217...<1.5$$ and we are done!
A: We can use also the following way.
Let $a=\sqrt{\frac{x}{y}}$ and $b=\sqrt{\frac{y}{z}}$, where $x$, $y$ and $z$ are positives.
Thus, $c=\sqrt{\frac{z}{x}}$ and by AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a}{a^a+1}\leq\sum_{cyc}\frac{a}{\frac{a^2+1}{2}+1}\leq\sum_{cyc}\frac{a}{2\sqrt{\frac{a^2+1}{2}\cdot1}}=\frac{1}{\sqrt2}\sum_{cyc}\sqrt{\frac{x}{x+y}}\leq$$
$$\leq\frac{1}{\sqrt2}\sqrt{\sum_{cyc}\frac{x}{(x+y)(x+z)}\sum_{cyc}(x+z)}=\sqrt{\frac{2(xy+xz+yz)(x+y+z)}{(x+y)(x+z)(y+z)}}.$$
Id est, it's enough to prove that
$$\frac{2(xy+xz+yz)(x+y+z)}{(x+y)(x+z)(y+z)}\leq\frac{9}{4}$$ or
$$\sum_{cyc}z(x-y)^2\geq0.$$
Done!
