Question

[Lyzimi]

cho \(a\ge0,b\ge0\)

cmr \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

Answer 1

[Trần Việt Anh]

Answer 2

[Thắng Nguyễn]

\(BDT\Leftrightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}-\frac{2}{1+ab}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\) đúng