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.
 A: 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).
A: 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.
A: (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
$$ \sum \frac{a^3}{a^2+b^2} \leq \frac{1}{4} \sum  \frac{ a^2+b^2}{a} \leq \sum \frac{1}{2}  \frac{ b^2}{a} $$
A: 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 $x\neq y$.
