<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] A funny solution (which seems correct): note that for $t>0$
$$
\frac{1}{1+t^2} \leq \frac{1}{2}t^2 + \frac{3}{2}(1-t).
$$
To prove this note first that $p(t) = (t^2 - 3t + 3)(t^2 + 1)$ has no real roots. Next, $p''(t) = 12t^2 -18t + 8 > 0$, implying that $p$ is strictly convex. Finally $p'(t) = 4t^3 - 9t^2 + 8t -3$, and $p'(1) = 0$ implying that $p$ has a minimum $2=p(1)\leq p(t)$.
Next, substitute $t\mapsto \frac{b}{a}$ and multiply both sides by $a$, then we get
$$
\frac{a^3}{a^2+b^2}\leq \frac{1}{2}\frac{b^2}{a} + \frac{3}{2}(a - b).
$$
Finally cyclically sum, and done.
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