<s>[INST] 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 !!
 [/INST] 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.
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