<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] Using Cauchy-Bunyakovsky-Schwarz,
\begin{align*}
  \frac{ab^2}{a^2 + b^2} + \frac{bc^2}{b^2 + c^2} + \frac{ca^2}{c^2 + a^2}
  \ge \frac{(b + c + a)^2}{\frac{a^2 + b^2}{a} + \frac{b^2 + c^2}{b} + \frac{c^2 + a^2}{c}}
  = \frac{(a + b + c)^2}{a + b + c + \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}}.
\end{align*}
It suffices to prove that
$$\frac{(a + b + c)^2}{a + b + c + \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}}
\ge a + b + c - \frac12\left(\frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}\right).$$
Denote $p = a + b + c$ and $Q = \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}$.
It suffices to prove that
$$\frac{p^2}{p + Q} \ge p - \frac12 Q$$
or
$$p^2 \ge (p + Q)(p - Q/2) = p^2 + pQ/2 - Q^2/2$$
or
$$Q \ge p$$
i.e.,
$$\frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c} \ge a + b + c$$
which is true using Cauchy-Bunyakovsky-Schwarz (easy).
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