Áp dụng BĐT AM-GM ta có:
\(a+b\ge2\sqrt{ab}\)
\(ab+1\ge2\sqrt{ab\cdot1}=2\sqrt{ab}\)
Nhân theo vế 2 BĐT ta có:
\(\left(a+b\right)\left(ab+1\right)\ge2\sqrt{ab}\cdot2\sqrt{ab}=4\sqrt{a^2b^2}=4ab\)
Đẳng thức xảy ra khi \(a=b\)
\(\left(a+b\right)\left(ab+1\right)\ge4ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(ab+1\right)}{ab}\ge4\)
\(\Leftrightarrow\left(\dfrac{a+b}{ab}\right)\left(ab+1\right)\ge4\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(ab+1\right)\ge4\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}\ge2\sqrt{\dfrac{1}{ab}}\\ab+1\ge2\sqrt{ab}\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(ab+1\right)\ge2\sqrt{\dfrac{1}{ab}}.2\sqrt{ab}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(ab+1\right)\ge4\) ( đpcm )
