` (a+b)( \frac{1}{a} + \frac{1}{b}) = 1 + 1 + \frac{a}{b} + \frac{b}{a} = 2 + \frac{a}{b} + \frac{b}{a} `
Áp dung BĐT cosi ta có
` \frac{a}{b} + \frac{b}{a} \geq ` $2\sqrt[]{\frac{a}{b} . \frac{b}{a}} = 2\sqrt[]{1} = 2$
Ta có
` (a+b)( \frac{1}{a} + \frac{1}{b}) = 1 + 1 + \frac{a}{b} + \frac{b}{a} = 2 + \frac{a}{b} + \frac{b}{a} `
mà ` \frac{a}{b} + \frac{b}{a} \geq 2 `
Nên ` (a+b)( \frac{1}{a} + \frac{1}{b}) \geq 2 +2=4`
=> đpcm
