Đề bài giả thiết phải cho \(x,y\le1\)nhé.
\(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\)
\(\Leftrightarrow\frac{1}{1+x^2}-\frac{1}{1+xy}\ge\frac{1}{1+xy}-\frac{1}{1+y^2}\)
\(\Leftrightarrow\frac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}\ge\frac{y^2-xy}{\left(1+xy\right)\left(1+y^2\right)}\)
\(\Leftrightarrow\frac{x\left(y-x\right)}{\left(1+xy\right)\left(1+x^2\right)}\ge\frac{y\left(y-x\right)}{\left(1+xy\right)\left(1+y^2\right)}\)
\(\Leftrightarrow\frac{\left(x-y\right)}{\left(1+xy\right)}\left(\frac{x}{1+x^2}-\frac{y}{1+y^2}\right)\ge0\)
\(\Leftrightarrow\frac{\left(x-y\right)}{\left(1+xy\right)}.\frac{x+xy^2-y-x^2y}{\left(1+x^2\right)\left(1+y^2\right)}\ge0\)
\(\Leftrightarrow\frac{\left(x-y\right)^2\left(1-xy\right)}{\left(1+xy\right)\left(1+x^2\right)\left(1+y^2\right)}\ge0\) (*)
Vì x,y \(\le1\) nên (*) luôn đúng.
Vậy bđt được chứng minh.
