How to prove that $\frac1a+\frac3b+2(a+\sqrt{b^2+8})>11$ Hello everyone: I want to prove that the following holds for all $a,b>0$: $$\frac1a+\frac3b+2(a+\sqrt{b^2+8})>11.$$
My attempt: We notice by AM-GM that $\frac1a+2a\geq 2\sqrt 2$ for all $a>0$. So we only need to prove that $$\frac3b+2\sqrt{b^2+8}>11-2\sqrt{2}.$$
This seems to be true numerically, but I don't know how to prove it.

Firstly, it should be $$\frac{3}{b}+4\sqrt{b^2+8}\geq11-2\sqrt2,$$ which is true by C-S and AM-GM:
$$\frac{3}{b}+4\sqrt{b^2+8}=\frac{3}{b}+\frac{4}{3}\sqrt{(1+8)(b^2+8)}\geq$$
$$\geq\frac{3}{b}+\frac{4(b+8)}{3}\geq4+\frac{32}{3}>11-2\sqrt2.$$
Secondly, we can prove your second inequality by the similar way:
$$\frac{3}{b}+2\sqrt{b^2+8}=\frac{3}{b}+\frac{2}{\sqrt{11}}\sqrt{(3+8)(b^2+8)}\geq$$
$$\geq\frac{3}{b}+\frac{2(\sqrt3b+8)}{\sqrt{11}}\geq2\sqrt{\frac{6\sqrt3}{\sqrt{11}}}+\frac{16}{\sqrt{11}}>11-2\sqrt2.$$
