Với a; b > 0
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}=\frac{4}{3}\)
\(ab\le\frac{\left(a+b\right)^2}{4}=\frac{9}{4}\)=> \(\frac{1}{ab}\ge\frac{4}{9}\)
Khi đó: \(S=\left(1+\frac{2}{a}\right)\left(1+\frac{2}{b}\right)=1+2\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{4}{ab}\ge1+2.\frac{4}{3}+4.\frac{4}{9}=\frac{49}{9}\)
Dấu "=" xảy ra <=> a = b = 3/2
vậy min S = 49/9