Let $a;b;c>0$. Prove : $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}+\frac{a^{2}}{a+b}$ Let $a;b;c>0$. Prove :
$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}+\frac{a^{2}}{a+b}$ 
P/s : Only use AM-GM and Cauchy-Schwarz
In the cases only use AM-GM and Cauchy-Schwarz; i don't have any thoughts about this problem !!
 A: I don't see how to do it with AM/GM and CS, but here's an alternative method: by homogeneity, we may assume wlog that $a+b+c=1$, so we are to prove that
$$ \frac{a^2}{1-a} + \frac{b^2}{1-b} + \frac{c^2}{1-c}
  \ge \frac{b^2}{1-a} + \frac{c^2}{1-b} + \frac{a^2}{1-c} $$
Since $x\mapsto x^2$ and $x\mapsto\frac1{1-x}$ are increasing functions for $x\in(0,1)$, the tuples $(a^2,b^2,c^2)$ and $(\frac1{1-a},\frac1{1-b},\frac1{1-c})$ are in the same order (e.g., if $a\ge b\ge c$ then $a^2\ge b^2\ge c^2$ and $\frac1{1-a}\ge\frac1{1-b}\ge\frac1{1-c}$), so the desired inequality follows from the rearrangement inequality.
A: We need to prove that $$\sum\limits_{cyc}\frac{a^2-b^2}{b+c}\geq0$$ or $$\sum\limits_{cyc}(a^2-b^2)(a+b)(a+c)\geq0$$ or
$$\sum\limits_{cyc}(a^2-b^2)(a^2+ab+ac+bc)\geq0$$ or
$$\sum\limits_{cyc}(a^2-b^2)a^2\geq0$$ or
$$\sum\limits_{cyc}\left(\frac{a^4+b^4}{2}-a^2b^2\right)\geq0,$$
which is true by AM-GM.
A: You can just do it in one step (without any computation) if you use the rearrangement inequality.
Notice that for any positive reals a, b and c, 
$(a^2,b^2,c^2)$ and $(1/b+c,1/c+a,1/a+c)$ are similarly sorted. So,

