How prove $\sum\limits_{cyc}(f(x^3)+f(xyz)-f(x^2y)-f(x^2z))\ge 0$ let $x,y,z\in (0,1)$, and the function
$$f(x)=\dfrac{1}{1-x}$$
show that
$$f(x^3)+f(y^3)+f(z^3)+3f(xyz)\ge f(x^2y)+f(xy^2)+f(y^2z)+f(yz^2)+f(z^2x)+f(zx^2)$$
For this problem simlar this Schur inequality:
http://www.artofproblemsolving.com/Wiki/index.php/Schur
we all know this third degree-Schur inequality,
$$a^3+b^3+c^3+3abc-ab(a+b)-bc(b+c)-ca(c+a)\ge 0$$
$a,b,c\ge0$
But my problem is $f(x)$ such this form.so How prove it?  Thank you
 A: In fact, this nice problem can indeed be solved using Schur's Inequality of 3rd degree, albeit indirectly. First we will prove a lemma:
Lemma: For $x, y, z>0$, we have:
$$e^{x^3}+e^{y^3}+e^{z^3}+3e^{xyz}\ge e^{x^2y}+e^{xy^2}+e^{y^2z}+e^{yz^2}+e^{z^2x}+e^{zx^2}$$
Proof: By Schur's inequality, for each $n\in\mathbb{N}$, we have:
$$G_n=\sum_{cyc} (\frac{x^n}{n!}-\frac{y^n}{n!})(\frac{x^n}{n!}-\frac{z^n}{n!})\ge 0 $$
Also, $\lim_{G_n\to\infty}=0$. Therefore:
$$\sum_{cyc}e^{x^3}+e^{xyz}-e^{x^2y}-e^{x^2z}=\sum_{n=1}^\infty G_n\ge 0$$
Where we here used $e^t=1+t+\frac{t^2}{2}+\cdots$. So the lemma is proven. $\Box$
Now take $(x, y, z):=(-x(\ln s)^{\frac{1}{3}}, -y(\ln s)^{\frac{1}{3}}, -z(\ln s)^{\frac{1}{3}})$ for a variable $s<1$. Then the lemma gives that:
$$g(s)=s^{-x^3}+s^{-y^3}+s^{-z^3}+3s^{-xyz}-( s^{-x^2y}+s^{-xy^2}+s^{-y^2z}+s^{-yz^2}+s^{-z^2x}+s^{-zx^2})\ge 0$$
for all $0\le s\le 1$ (the endpoints are true by inspection). Finally, integrating this over the interval $[0, 1]$ gives:
$$\int_0^1 g(s)=\sum_{cyc}\frac{1}{1-x^3}+\frac{1}{1-xyz}-\frac{1}{1-x^2y}-\frac{1}{1-x^2z}\ge 0$$
as desired. 
