Question

cho \(a\ge2\) tìm min: \(a+\dfrac{1}{a}\)

Answer 1

\(a+\dfrac{1}{a}=\dfrac{a}{4}+\dfrac{1}{a}+\dfrac{3a}{4}\)

Theo AM - GM thì

\(\dfrac{a}{4}+\dfrac{1}{a}\ge1\) \(\Rightarrow\dfrac{a}{4}+\dfrac{1}{a}+\dfrac{3a}{4}\ge1+\dfrac{3a}{4}\ge1+\dfrac{6}{4}=\dfrac{5}{2}\)

min = 5/2 khi a = 2

Answer 2

BĐT Cosi :

\(a+\dfrac{1}{a}\ge2\sqrt{a.\dfrac{1}{a}}=2\)