<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] We can prove it without expending by SOS and the Tangent Line method.
Indeed, $$\sum_{cyc}\frac{b^2-a^2}{a+c}=\frac{1}{2}\sum_{cyc}\frac{2b^2-2a^2}{a+c}=\frac{1}{2}\sum_{cyc}\left(\frac{2b^2-2a^2}{a+c}+a-c\right)=$$
$$=\frac{1}{2}\sum_{cyc}\frac{b^2-c^2-(a^2-b^2)}{a+c}=\frac{1}{2}\sum_{cyc}\left(\frac{a^2-b^2}{c+b}-\frac{a^2-b^2}{a+c}\right)=\sum_{cyc}\frac{(a-b)^2(a+b)}{2(a+c)(b+c)}\geq0.$$
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