=1+1/a+1/b+1/ab (1)
Áp dụng Cosy ta có 1/a+1/b>=4/(a+b)=4 (2)
(a+b)^2>=4ab nên ab<=(a+b)^2/4=1/4 hay 1/ab>=4 (3)
Từ (1)(2)(3) ta đc 1+1/a+1/b+1/ab>=1+4+4=9 (đpcm)
Ta có: \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)=\left(1+\frac{a+b}{a}\right)\left(1+\frac{a+b}{b}\right)\) \(=\left(1+1+\frac{b}{a}\right)\left(1+1+\frac{a}{b}\right)\) \(=\left(2+\frac{b}{a}\right)\left(2+\frac{a}{b}\right)\) \(=4+2\left(\frac{a}{b}+\frac{b}{a}\right)+\frac{ab}{ab}\) \(=5+2\left(\frac{a}{b}+\frac{b}{a}\right)\)
. Áp dụng BĐT Cô-si cho 2 số \(\frac{a}{b}\) và \(\frac{b}{a}\) , ta có:
\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{ab}{ab}}=2\) . Suy ra \(2\left(\frac{a}{b}+\frac{b}{a}\right)\ge4\)
. Suy ra \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge5+4=9\) (đpcm)
. Dấu "=" xảy ra khi \(a=b\)
