S=$4+\frac{2}{a}+\frac{2}{b}+a+b+\frac{a}{b}+\frac{b}{a}=4+2(\frac{1}{a}+\frac{1}{b})+a+b+\frac{a}{b}+\frac{b}{a}$ $\geq 4+\frac{8}{a+b}+a+b+2$
ta có $(a+b)^{2}\leq 2(a^{2}+b^{2})=2 \Rightarrow a+b\leq \sqrt{2}$
đặt $t=a+b \leq \sqrt{2}$
$S\geq 6+\frac{8}{t}+t=(t+\frac{2}{t})+\frac{6}{t}+6\geq 6+5\sqrt{2}$
Dấu "=" $\Leftrightarrow a=b=\frac{1}{\sqrt{2}}$