Question

CMR : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{ab}\) (a,b >0)

Answer 1

Áp dụng BĐT Cô - Si vào bài toán , ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}\) ≥ \(\dfrac{2}{\sqrt{ab}}\)

⇔\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) ≥ \(\dfrac{2}{\sqrt{ab}}.2\sqrt{ab}\)

⇔\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) ≥ 4

⇔ \(\dfrac{1}{a}+\dfrac{1}{b}\) ≥ \(\dfrac{4}{a+b}\)

Dấu " = " xảy ra khi : a = b

Answer 2

Áp dụng BĐT Cô - si dạng Engel , ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}\) ≥ \(\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{a+b}\)

Dấu " = " xảy ra khi : a = b