Given $a, b, c>0$, prove $\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b)$ 
Given $a,b,c>0$, prove that $$\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b).$$

My attempt:
I have that $$\frac{a^4}{a+b}+\frac{c^2(a+b)}{4}\geq a^{2}c$$
$$\frac{b^4}{b+c}+\frac{a^2(b+c)}{4}\geq b^{2}a$$
$$\frac{c^4}{c+a}+\frac{b^2(c+a)}{4}\geq c^{2}b$$
Adding them up yields $$\sum \frac{a^4}{a+b}\geq\frac{3}{4}\sum a^2c-\frac{1}{4}\sum a^2b.$$
So it boils down to proving that $a^2b+b^2c+c^2a\leq ab^2+bc^2+ca^2$, which isn't correct.
Could you help me with this problem?

We need to prove that: $$\sum_{cyc}\left(\frac{a^4}{a+b}-\frac{1}{2}ab^2\right)\geq0$$
or
$$\sum_{cyc}\frac{a(2a^3-ab^2-b^3)}{a+b}\geq0$$ or
$$\sum_{cyc}\frac{a(a-b)(2a^2+2ab+b^2)}{a+b}\geq0$$ or
$$\sum_{cyc}\left(\frac{a(a-b)(2a^2+2ab+b^2)}{a+b}-\frac{5}{6}(a^3-b^3)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2(7a^2+9ab+5b^2)}{a+b}\geq0$$ and we are done!
Now we see that our inequality is true even for any $a$, $b$ and $c$ 
such that $a+b>0$, $a+c>0$ and $b+c>0$.
