The bump

Quidditch wrote: Let $a, b, c \in\mathbb{R}^+$. Prove that $$\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)<\sum_{cyc}\frac{a^3+b^3}{c^2+ab}.$$ Let $a\geq b\geq c$. Thus, by Holder we obtain: $$\sum_{cyc}\frac{a^3+b^3}{c^2+ab}-\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)\geq\frac{2(a+b+c)^3}{3\sum\limits_{cyc}(a^2+ab)}-\sum_{cyc}\frac{2ab(a+b+c)}{(2a+c)(2b+c)}=$$$$=2(a+b+c)\left(\frac{(a+b+c)^2}{3\sum\limits_{cyc}(a^2+bc)}-\frac{1}{3}+\sum_{cyc}\left(\frac{ab}{3(ab+ac+bc)}-\frac{ab}{(2a+c)(2b+c)}\right)\right)\geq$$$$\geq\frac{2(a+b+c)}{3(ab+ac+bc)}\sum_{cyc}\frac{ab(c-a)(c-b)}{(2a+c)(2b+c)}\geq\frac{2(a+b+c)}{3(ab+ac+bc)}\left(\frac{ab(c-a)(c-b)}{(2a+c)(2b+c)}+\frac{ac(b-a)(b-c)}{(2a+b)(2c+b)}\right)=$$$$=\frac{2(a+b+c)a(b-c)^2(2(b+c)a^2+(b^2+5bc+c^2)a-bc(b+c))}{3(ab+ac+bc)(2a+b)(2c+b)(2a+c)(2b+c)}\geq0.$$Another way based on the following C-S: $$\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)\leq\frac{1}{9}\sum_{cyc}ab\left(\frac{1}{a}+\frac{4}{a+c}+\frac{1}{b}+\frac{4}{b+c}\right)=\frac{2}{3}(a+b+c).$$