\(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\)
\(\frac{1+y^2+1+x^2}{\left(1+x^2\right)\left(1+y^2\right)}\ge\frac{2}{1+xy}\\ \frac{2+x^2+y^2}{\left(1+x^2\right)\left(1+y^2\right)}\ge\frac{2}{1+xy}\)
=>\(\left(2+x^2+y^2\right)\left(1+xy\right)\ge2\left(1+x^2\right)\left(1+y^2\right)\)
\(\left(2+x^2+y^2\right)+\left(2+x^2+y^2\right)xy\ge2\left(1+x^2\right)\left(1+y^2\right)\)
\(2+x^2+y^2+2xy+x^3y+y^3x\ge\left(2+2x^2\right)\left(1+y^2\right)\)
\(2+x^2+y^2+2xy+x^3y+y^3x\ge2+2x^2+\left(2+2x^2\right)y^2\)
\(2+x^2+y^2+2xy+x^3y+y^3x\ge2+2x^2+2y^2+2x^2y^2\)
\(2xy+x^3y+y^3x\ge x^2+y^2+2x^2y^2\)
\(2xy+x^3y+y^3x-x^2-y^2-2x^2y^2\ge0\)
\(x^3y-x^2+y^3x-y^2+2xy-2x^2y^2\ge0\)
\(x^2\left(xy-1\right)+y^2\left(xy-1\right)-2xy\left(xy-1\right)\)\(\ge0\)
\(\left(xy-1\right)\left(x^2-2xy+y^2\right)\ge0\)
\(\left(xy-1\right)\left(x-y\right)^2\ge0\)
\(Do\begin{cases}x,y\ge1=>xy\ge1=>xy-1\ge0\\\left(x-y\right)^2\ge0\end{cases}\)
\(=>\left(xy-1\right)\left(x-y\right)^2\ge0\left(dpcm\right)\)
