<s>[INST] Inequality in cyclic order : $\sum\frac{8}{(a+b)^2+4abc}+a^2+b^2+c^2\ge\sum\frac{8}{a+3}$ 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.
 [/INST] 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,
$$\sum_{cyc}\left( \frac{4}{(a^2+b^2)(c+1)}+\frac{a^2+b^2}{2}\right)\ge \sum_{cyc} \frac{8}{a+3}$$

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


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

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