Question

Cho\(x;y\ge1\). Chứng minh rằng:

\(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\)

Answer 1

Ta có: \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\)

\(\Leftrightarrow\left(\frac{1}{1+x^2}-\frac{1}{1+xy}\right)+\left(\frac{1}{1+y^2}-\frac{1}{xy}\right)\ge0\)

\(\Leftrightarrow\frac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\frac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow x\left(y-x\right)\left(1+y^2\right)+y\left(x-y\right)\left(1+x^2\right)\ge0\)

\(\Leftrightarrow\left(y-x\right)\left(x+xy^2-y-x^2y\right)\ge0\)

\(\Leftrightarrow\left(y-x\right)^2\left(xy-1\right)\ge0\)(đúng với mọi x,y>=1)