Áp dụng BĐT Bunhiacopxki:
\(\left(1+ab\right)\left(1+\dfrac{a}{b}\right)\ge\left(1+a\right)^2\)
\(\Rightarrow\dfrac{1}{\left(1+a\right)^2}\ge\dfrac{1}{\left(1+ab\right)\left(1+\dfrac{a}{b}\right)}=\dfrac{b}{\left(a+b\right)\left(1+ab\right)}\)
Tương tự:
\(\dfrac{1}{\left(1+b\right)^2}\ge\dfrac{a}{\left(a+b\right)\left(1+ab\right)}\)
Cộng vế:
\(\dfrac{1}{\left(1+a\right)^2}+\dfrac{1}{\left(1+b\right)^2}\ge\dfrac{a+b}{\left(a+b\right)\left(1+ab\right)}=\dfrac{1}{1+ab}\)
Dấu "=" xảy ra khi \(a=b=1\)