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

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 $b+c>0$.