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.

Let me continue your approach. We will use the following simple observation:
Lemma. For positive $x$ and $y$ the following inequality holds
$$
\frac{x^2}{y}\ge 2x-y.
$$
Proof. It is equivalent to $(x-y)^2\ge 0$.
We need to prove that
$$
\sum_{cyc}\frac{(ab)^2}{a^3+ab^2}\ge \frac{1}{2}\sum_{cyc}\frac{2a^2-b^2}{a}.
$$
Note that
$$
\frac{(ab)^2}{a^3+ab^2}=\frac{b^2}{a+\frac{b^2}{a}}=\frac{1}{4}\cdot\frac{(2b)^2}{a+\frac{b^2}{a}}\ge\frac{1}{4}\left(4b-a-\frac{b^2}{a}\right).
$$
Thus, summing up similar inequalities we obtain
$$
\sum_{cyc}\frac{(ab)^2}{a^3+ab^2}\ge\frac{3}{4}(a+b+c)-\frac{1}{4}\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).
$$
It remains to check that
$$
\frac{3}{4}(a+b+c)-\frac{1}{4}\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)\ge\frac{1}{2}\sum_{cyc}\frac{2a^2-b^2}{a}.
$$
Since $\sum_{cyc}\frac{2a^2-b^2}{a}=2(a+b+c)-\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)$, the last inequality is equivalent to
$$
\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\ge a+b+c,
$$
which is again a consequence of the lemma:
$$
\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\ge (2b-a)+(2c-b)+(2a-c)=a+b+c.
$$
Comment. Here is a more general form of the lemma: $\frac{a^x}{b^y}\ge\frac{xa^{x-y}-yb^{x-y}}{x-y}$ for any $a,b,x,y>0$ such that $x\neq y$.
