<s>[INST] $\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof Does anyone know hot to prove this inequality?
Having: $a, b, c \gt 0$
$$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$
I tried with the AM-GM inequality but I couldn't get any improvement. I'm on a still point and I don't know how to continue.
Furthermore, I don't know how to get rid of the known term. I tried with the AM-HM inequality but I'm still not getting any results.
Also, just some hints would be appreciated, thanks
 [/INST] Since the expression is cyclic, we can, WLOG, reduce it into two cases:
Suppose $a\ge b\ge c > 0$. We have
$$\frac{a^2-c^2}{b+c} = \frac{a^2-b^2}{c+b} + \frac{b^2-c^2}{c+b} \ge \frac{a^2-b^2}{c+a} + \frac{b^2-c^2}{a+b}$$
Now suppose $a \ge c \ge b > 0$. We then have
$$\frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge \frac{c^2-b^2}{a+c} + \frac{a^2-c^2}{a+c} =\frac{a^2-b^2}{c+a}$$
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