Inequality : $\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)<\sum_{cyc}\frac{a^3+b^3}{c^2+ab}$ 
Let $\{a, b, c\} \subset \mathbb{R}^+$. Prove that 
  $$\displaystyle\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)<\displaystyle\sum_{cyc}\frac{a^3+b^3}{c^2+ab}$$


My work :
WLOG, let  $\;c\geq b \geq a$, we have
$$\frac{b^3+c^3}{a^2+bc}\geq \frac{b^3+c^3}{b^2+bc}$$
Since $(b-c)(b-a) \leq 0$, we have
$$\frac{c^3+a^3}{b^2+ac}\geq \frac{c^3+a^3}{bc+ba}. $$
$\begin{eqnarray}
\sum_{cyc}\frac{a^3+b^3}{c^2+ab}&\geq &\frac{a^3+b^3}{c^2+ab} + \frac{b^3+c^3}{b^2+bc} + \frac{c^3+a^3}{bc+ba} \\
&\geq& \frac{a^3+b^3}{c^2+ab} + \frac{b^2-bc+c^2}{b} + \frac{c^2-ca+a^2}{c}\\
&\geq &\frac{a^3+b^3}{c^2+ab} + b-a+\frac{c^2}{b} + \frac{a^2}{c}=S
\end{eqnarray}$

Case 1 : $b-a \geq c-b$
we have $ \frac{b}{a} \geq \frac{c}{b}$, so $b^2\geq ac, \;2b \geq a+c$
so  $$ S \geq \frac{a^3+b^3}{c^2+ab} +c-b+\frac{c^2}{b} + \frac{a^2}{c}$$
$$\geq \left(\frac{c^2}{b}+c\right)+\left(\frac{a^3+b^3}{c^2+ab}-b+  \frac{a^2}{c}\right)$$
Since $\frac{2}{9}\left(\frac{bc}{a}\right)=\frac{2}{9}\left(\frac{c^2a+b^2c-c^2a}{ab}\right)=\frac{2}{9}\left(\frac{c^2}{b}\right)+\frac{2}{9}\left(\frac{c(b^2-ca)}{ab}\right)\geq \frac{2}{9}\left(\frac{c^2}{b}\right)$
As $\frac{4}{9}a + \frac{4}{9}b + \frac{4}{9}c < \frac{8}{9}b + \frac{4}{9}\frac{c^2}{b}$, so $\frac{2}{9}\frac{ab}{c} < \frac{2}{9}\frac{abc}{c^2}< \frac{2}{9}c$
Similarly, $\frac{2}{9}\frac{ac}{b} < \frac{2}{9}c$
Thus, $\frac{4}{9}(a+b+c)+\frac{2}{9}\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right) < S $

Case 2 : $b-a \leq c-b$
we have $b^2\leq ac, \;2b \leq a+c$
so $ S \geq \left(\frac{c^2}{b}+c\right)+\left(\frac{a^3+b^3}{c^2+ab}-b+  \frac{a^2}{c}\right)\geq \left(\frac{2c^2}{a+c}+c\right)+\left(\frac{a^3+b^3}{c^2+ab}-b+  \frac{a^2}{c}\right)$
Since $b^2 \leq ac$, so  $2ca \geq b(a+c)$ 
$\leftrightarrow c(2ca) \geq c(ab)+c(bc)$
$\leftrightarrow  \frac{2c^2}{a+c} \geq \frac{bc}{a}$
Thus, $\frac{2c^2}{a+c}+c \geq \frac{bc}{a}+c > \frac{4}{9}(a+b+c) +\frac{2}{9}\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)$

By C-S,
$\frac{1}{2a+b}\leq \frac{2}{a}+ \frac{1}{b}$
Similarly, $\frac{1}{2b+c}\leq \frac{2}{b}+ \frac{1}{c}$, so
$\displaystyle\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)\leq \frac{4}{9}(a+b+c) +\frac{2}{9}\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)$
therefore, 
$\displaystyle\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)<\displaystyle\sum_{cyc}\frac{a^3+b^3}{c^2+ab}$  $\blacksquare$



*

*Please help me check if there is any mistake in my work. Thank you.

*If you have different idea, please provide.
 A: By C-S
$$\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)=\sum_{cyc}ab\left(\frac{1}{a+a+c}+\frac{1}{b+b+c}\right)\leq$$
$$\leq\sum_{cyc}\frac{ab}{9}\left(\frac{1^2}{a}+\frac{2^2}{a+c}+\frac{1^2}{b}+\frac{2^2}{b+c}\right)=$$
$$=\frac{2}{9}(a+b+c)+\frac{4}{9}\sum_{cyc}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=$$
$$=\frac{2}{9}(a+b+c)+\frac{4}{9}\sum_{cyc}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)=\frac{2}{9}(a+b+c)+\frac{4}{9}(a+b+c)=\frac{2}{3}(a+b+c).$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{a^3+b^3}{c^2+ab}\geq\frac{2}{3}(a+b+c).$$
We'll prove a stronger inequality:
$$\sum_{cyc}\frac{a^3+b^3}{c^2+ab}\geq a+b+c.$$
Indeed, by C-S
$$\sum_{cyc}\frac{a^3+b^3}{c^2+ab}=\sum_{cyc}\frac{a^4}{c^2a+a^2b}+\sum_{cyc}\frac{b^4}{c^2b+b^2a}\geq$$
$$\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(c^2a+a^2b)}+\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(c^2b+b^2a)}=$$
$$=\frac{(a^2+b^2+c^2)^2}{2}\left(\frac{1}{a^2b+b^2c+c^2a}+\frac{1}{a^2c+b^2a+c^2b}\right)\geq$$
$$\geq\frac{(a^2+b^2+c^2)^2}{2}\cdot\frac{4}{\sum\limits_{cyc}(a^2b+a^2c)}=\frac{2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b+a^2c)}.$$
Thus, it remains to prove that
$$2(a^2+b^2+c^2)^2\geq(a+b+c)\sum_{cyc}(a^2b+a^2c)$$ or
$$\sum_{cyc}(2a^4-a^3b-a^3c+2a^2b^2-2a^2bc)\geq0,$$
which is true by Muirhead.
Done!
