Question

Với a,b > 0, chứng minh :
\(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\). Dấu " = " xảy ra khi nào ?

Answer 1

Ta có : \(\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{1}{4}.\dfrac{4}{a+b}=\dfrac{1}{a+b}\) ( đpcm )

Dấu \("="\) xảy ra khi \(a=b\)

Answer 2

\(bđt\Leftrightarrow\dfrac{1}{a+b}\le\dfrac{a+b}{4ab}\)

\(\Leftrightarrow4ab\le\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\left(luôn-đúng\right)\)

\("="\Leftrightarrow a=b\)