A property of Nesbitt's inequality I think it's not new however I think  it's interesting  :

Let $a,b,c>0$ and $x>0$ then the function :
  $$f(x)=\frac{a^x}{b^x+c^x}+\frac{b^x}{a^x+c^x}+\frac{c^x}{b^x+a^x}$$
$f(x)$ is increasing

My first try :
The derivative of $f(x)$ is :
$$f'(x)=\sum_{cyc}\frac{a^x\ln(a)(b^x+c^x)-a^x(b^x\ln(b)+c^x\ln(c))}{(b^x+c^x)^2}$$
We multiply by $x$ it gives :
$$f'(x)=\frac{1}{x}\Big(\sum_{cyc}\frac{a^x\ln(a^x)(b^x+c^x)-a^x(b^x\ln(b^x)+c^x\ln(c^x))}{(b^x+c^x)^2}\Big)$$
$x$ is positive so we have to prove :
$$\sum_{cyc}\frac{a^x\ln(a^x)(b^x+c^x)-a^x(b^x\ln(b^x)+c^x\ln(c^x))}{(b^x+c^x)^2}\geq 0$$
We put $a^x=u$ , $b^x=v$ and $c^x=w$ so the inequality is equivalent to :
$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq 0$$
Now with the concavity of $\ln(x)$ and the inequality of Jensen's we have :
$$\ln\Big(\frac{v^2+w^2}{v+w}\Big)\geq \frac{v\ln(v)+w\ln(w)}{v+w}$$
So we have :
$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq\sum_{cyc}\frac{u\ln(u)-u\Big(\ln\Big(\frac{v^2+w^2}{v+w}\Big)\Big)}{(v+w)}\geq 0$$
Or :
$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq\sum_{cyc}\frac{u\Big(\ln\Big(\frac{u(v+w)}{v^2+w^2}\Big)\Big)}{(v+w)}\geq 0$$
Now we have to prove :
$$\sum_{cyc}\frac{u\Big(\ln\Big(\frac{u(v+w)}{v^2+w^2}\Big)\Big)}{(v+w)}\geq 0$$
Unfortunatly I'm stuck here ...
Second try :
For $a,b,c>0$ the function $g(x)=\frac{a^x}{b^x+c^x}$ is convex so the function $f(x)$ is convex as sum of convex functions .
Now we have with $a,b,c>0$ :
$$\sum_{cyc}\frac{a^2}{b^2+c^2}\geq \sum_{cyc}\frac{a}{b+c}$$
Fact wich have been proven by Michael Rozenberg (see my questions)
Remains to apply the three chord lemma to get the increase of the function $f(x)$
My question :
Can someone could find an alternative proof or prove this last inequality ?
There are generalizations of this facts ? 

Alternative solution:
WLOG, assume $a \ge b \ge c = 1$. Denote $a^x = u, b^x = v$. Then $u \ge v \ge 1$.
We have
\begin{align*}
f'(x) &= A \ln a + B\ln b \\
&= A\ln \frac{a}{b} + (A + B) \ln b
\end{align*}
where
\begin{align*}
A &= {\frac {u}{v+1}} - {\frac {vu}{ \left( 1+u \right) ^{2}}} -{\frac {u}{
 \left( u+v \right) ^{2}}},\\
B &={\frac {v}{1+u} -{\frac {vu}{ \left( v+1
 \right) ^{2}}}} -{\frac {v}{ \left( u+v \right) ^{2}}}. 
\end{align*}
We have
$$A \ge {\frac {u}{v+1}} - {\frac {vu}{ \left( 1+v \right) ^{2}}} -{\frac {u}{
 \left( 1+v \right) ^{2}}} = 0.$$
Also, we have
$$A + B = \frac{u}{(1 + v)^2} - \frac{u}{(u + v)^2} + \frac{v}{(1 + u)^2} - \frac{v}{(u + v)^2} \ge 0.
$$
Thus, $f'(x) \ge 0$ for all $x > 0$.
We are done.
