Inequality in cyclic order : $\sum\frac{8}{(a+b)^2+4abc}+a^2+b^2+c^2\ge\sum\frac{8}{a+3}$ [closed]

Prove:$$\sum_{cyc} \frac{8}{(a+b)^2+4abc}+a^2+b^2+c^2\ge\sum_{cyc} \frac{8}{a+3}$$ where $a,~b,~c$ are positive real numbers.

My thought: I think Holder's inequality will be used. But can't understand how to start? Please give any hint.

Solution 1:

Hint: Using A.M-G.M inequality, $$(a+b)^2\le2(a^2+b^2)~~\text{and}~~ 4abc\le2c(a^2+b^2) $$ $$\therefore\; (a+b)^2+4abc\le 2(a^2+b^2)(c+1) $$ It remains to prove, $$\sum_{cyc}\left( \frac{4}{(a^2+b^2)(c+1)}+\frac{a^2+b^2}{2}\right)\ge \sum_{cyc} \frac{8}{a+3}$$

$$ \frac{4}{(a^2+b^2)(c+1)}+\frac{a^2+b^2}{2}\ge {\frac{2\sqrt{2}}{\sqrt{c+1}}}$$

$$\frac{2+(a+1)}{2}\ge \sqrt{2(a+1)} \implies \frac{4}{\sqrt{2(a+1)}}\ge\frac{8}{a+3}$$