<s>[INST] Prove that $\sum \frac{a^3}{a^2+b^2}\le \frac12 \sum \frac{b^2}{a}$ 
Let $a,b,c>0$. Prove that
$$ \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\le \frac12 \left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).\tag{1}$$

A idea is to cancel the denominators, but in this case Muirhead don't work because the inequality is only cyclic, not symmetric.
Another idea would be to apply Cauchy reverse technique:
$$(1)\iff\sum \left(a-\frac{a^3}{a^2+b^2}\right)\ge \frac12 \sum (2a-b^2/a)\iff \sum\frac{ab^2}{a^2+b^2}\ge \frac12\sum\frac{2a^2-b^2}{a}$$
$$\iff \sum \frac{(ab)^2}{a^3+ab^2}\ge \frac12\sum \frac{2a^2-b^2}a.$$
Now we can apply Cauchy-Schwarz, and the problem reduces to
$$\frac{(\sum ab)^2}{\sum a^3+\sum ab^2}\ge \frac12\sum \frac{2a^2-b^2}{a},$$
and at this point I am stuck. Here the only idea is to cancel the denominators, but as I say above it can't work.
 [/INST] (Just AM-GM is sufficient.)
From OP's work / River Li's solution, the stated inequality is equivalent to $\sum a - \frac{a^3}{a^2+b^2} \geq \sum a - \frac{1}{2} \sum \frac{b^2}{a}$, which is:
$$ \sum \frac{ab^2 } { a^2 + b^2 } \geq \sum a - \frac{1}{2} \sum \frac{ b^2}{a}$$
Let $ X = \sum \frac{ab^2 } { a^2 + b^2 } , Y = \sum a, Z = \sum \frac{ b^2}{a}$ for simplicity.
We want to show that $  X \geq Y  - \frac{1}{2} Z$.
Notice that by AM-GM,
$$ \frac{ 4ab^2 }{ a^2 + b^2 } + \frac{ a^2 + b^2 } { a} \geq 4b, 
\quad \frac{ b^2}{a} + a \geq 2b. $$
Summing up the cyclic versions gives us

*

*$4X + Y + Z \geq 4Y$

*$ Z + Y \geq 2Y \Rightarrow Z \geq Y$.

*Hence $4 X \geq 3Y - Z \geq 4Y - 2Z$ as desired.


Note: From $ 4X \geq 4Y - (Y+Z)$, the inequality could be strengthened to
$$ \sum \frac{a^3}{a^2+b^2} \leq \frac{1}{4} \sum  \frac{ a^2+b^2}{a} \leq \sum \frac{1}{2}  \frac{ b^2}{a} $$
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