Ta có:
\(\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{1}{4a}+\dfrac{1}{4b}\)
Áp dụng bất đẳng thức AM-GM ta có:
\(\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{2\sqrt{ab}}\\\dfrac{1}{4a}+\dfrac{1}{4b}\ge\dfrac{1}{2\sqrt{16ab}}=\dfrac{1}{2.4\sqrt{ab}}=\dfrac{1}{8\sqrt{ab}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Chúc bạn học tốt!!!