For $a>0,b>0,c>0$, prove $\frac{a(a^2+bc)}{b+c}+\frac{b(b^2+ca)}{bc+a}+\frac{c(c^2+ab)}{a+b}\geq ab+bc+ca$

Question

Suppose $a>0,b>0,c>0$, prove $$\frac{a(a^2+bc)}{b+c}+\frac{b(b^2+ca)}{c+a}+\frac{c(c^2+ab)}{a+b}\geq ab+bc+ca.$$ Any help will welcome! This problem is from a competition. I just hope someone can give me some hints to start this problem! I begin this with $$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq \frac{ab+bc+ca}{2}.$$ But it seems no use for this problem.

Answer 1

It is equivalent to $$\frac{(a+b+c)(a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2)}{(b+c)(c+a)(a+b)}\geq 0$$ Now just use that $$x^2+y^2+z^2\geq xy+yz+zx$$ for all real $x,y,z$ is hold.

Answer 2

$$\sum_{cyc}\frac{a^3+abc}{b+c}-ab-ac-bc=\sum_{cyc}\left(\frac{a^3+abc}{b+c}-\frac{a(b+c)}{2}\right)=$$ $$=\sum_{cyc}\frac{a(2a^2-b^2-c^2)}{2(b+c)}=\sum_{cyc}\frac{a((a-b)(a+b)-(c-a)(a+c))}{2(b+c)}=$$ $$=\sum_{cyc}(a-b)\left(\frac{a(a+b)}{2(b+c)}-\frac{b(a+b)}{2(a+c)}\right)=\sum_{cyc}\frac{(a-b)^2(a+b)(a+b+c)}{2(a+c)(b+c)}\geq0.$$