<s>[INST] 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?
 [/INST] Similar to your inequalities, we have  
$$ \frac{ 3a^4}{a+b} +  \frac{3}{4}c^2 (a+b) \geq 3a^2c $$
$$ \frac{ a^4}{a+b} +   \frac{1}{4} b^2(a+b) \geq a^2b $$
$$ \frac{ a^4}{a+b} +   \frac{1}{4} a^2(a+b) \geq a^3 $$
Taking the cyclic summation, we get   
$$ \sum \frac{ 5a^4 } { a+b} \geq  \sum  2a^2 c  + \sum \frac{1}{2}a^3 $$
By rearrangement, we know that $ \sum \frac{1}{2}a^3  \geq \sum\frac{1}{2} a^2 c$, thus,
$$ \sum \frac{ a^4 } { a+b}  \geq \sum \frac{1}{2} a^2c, $$

The second inequality balances out the $a^2 b$ term which OP had (and is why we have a $3:1$ ratio).
The third inequality balances out the $a^3$ term which the second inequality introduces, and thankfully the other term from the third inequality cancels out with the first.   
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