# 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,

$\displaystyle\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)<\displaystyle\sum_{cyc}\frac{a^3+b^3}{c^2+ab}$ $\blacksquare$

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.
• @carat I checked your proof. I think it's wrong because the inequality $\sum\limits_{cyc}\left(\frac{2}{9}a+\frac{4}{9}\frac{ab}{c}\right)<\sum\limits_{cyc}\frac{a^3+b^3}{b^2+ac}$ is wrong. Try $c\rightarrow0^+$. Sep 14, 2017 at 4:11