$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ 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
 A: 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
$$\frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge \frac{c^2-b^2}{a+c} + \frac{a^2-c^2}{a+c} =\frac{a^2-b^2}{c+a}$$
A: I am assuming $a,b,c>0$. Clear denominators, expand, cancel like terms and obtain
$$a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2\geq 0$$
However, this is easily proved by cyclically adding up the obvious inequalities $\frac{1}{2}a^4+\frac{1}{2}b^4\geq a^2b^2$.
A: We can prove it without expending by SOS and the Tangent Line method.
Indeed, $$\sum_{cyc}\frac{b^2-a^2}{a+c}=\frac{1}{2}\sum_{cyc}\frac{2b^2-2a^2}{a+c}=\frac{1}{2}\sum_{cyc}\left(\frac{2b^2-2a^2}{a+c}+a-c\right)=$$
$$=\frac{1}{2}\sum_{cyc}\frac{b^2-c^2-(a^2-b^2)}{a+c}=\frac{1}{2}\sum_{cyc}\left(\frac{a^2-b^2}{c+b}-\frac{a^2-b^2}{a+c}\right)=\sum_{cyc}\frac{(a-b)^2(a+b)}{2(a+c)(b+c)}\geq0.$$
A: Because
$$\sum \frac{a^2}{c+a} = \sum \frac{c^2}{c+a}.$$
Therefore, we need to prove
$$2\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right) \geqslant \frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}.$$
Suppose $a \geqslant b \geqslant c,$ then
$$a^2 \geqslant b^2 \geqslant c^2 \quad \text{and} \quad \frac{1}{b+c} \geqslant \frac{1}{c+a} \geqslant \frac{1}{a+b}.$$
Using the Rearrangement inequality
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geqslant \frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}.$$
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geqslant \frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}.$$
The proof is completed.
A: Use general Schur's inequality to get (if $a\geqslant b\geqslant c$)
$$
\sum_{cyc}{\frac{(b+a)(b-a)}{c+a}} = \sum_{cyc}{\frac{(b-a)(b-c)}{c+a}} \geqslant 0
$$
For the other case (assuming $c$ is minimal), make substitution
$p$ for $b$,
$q$ for $a$ and
$r$ for $c$. Rest expressions are the same.
