<s>[INST] 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 ? 
 [/INST] By your assumption 
$$f'(x)=\sum_{cyc}\frac{a^x\ln{a}(b^x+c^x)-a^x(b^x\ln{b}+c^x\ln{c})}{(b^x+c^x)^2}=$$
$$=\sum_{cyc}\frac{a^x(b^x(\ln{a}-\ln{b})-c^x(\ln{c}-\ln{a}))}{(b^x+c^x)^2}=$$
$$=\sum_{cyc}(\ln{a}-\ln{b})\left(\frac{a^xb^x}{(b^x+c^2)^2}-\frac{a^xb^x}{(c^x+a^x)^2}\right)=$$
$$=\sum_{cyc}\frac{a^xb^x(\ln{a}-\ln{b})(a^x-b^x)(a^x+b^x+2c^x)}{(b^x+c^x)^2(c^x+a^x)^2}\geq0$$ because $$(\ln{a}-\ln{b})(a^x-b^x)=b^x\ln\frac{a}{b}\left(\left(\frac{a}{b}\right)^x-1\right)\geq0$$ for all $x>0$ and we are done! 
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