Cho hai số thực x, y là các số lớn hơn hoặc bằng 1. Chứng minh rằng: $\frac{1}{1+x^{2}}$ +$\frac{1}{1+y^{2}}$ $\geq$ $\frac{2}{1+xy}$ 

Giải thích các bước giải:

Ta có:

$\dfrac{1}{1+x^2}+\dfrac1{1+y^2}-\dfrac{2}{1+xy}$

$=\left(\dfrac1{1+x^2}-\dfrac1{1+xy}\right)+\left(\dfrac1{1+y^2}-\dfrac1{1+xy}\right)$

$=\dfrac{\left(1+xy\right)-\left(1+x^2\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{\left(1+xy\right)-\left(1+y^2\right)}{\left(1+y^2\right)\left(1+xy\right)}$

$=\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}$

$=-\dfrac{x\left(x-y\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}$

$=\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}-\dfrac{x\left(x-y\right)}{\left(1+x^2\right)\left(1+xy\right)}$

$=\dfrac{y\left(x-y\right)\left(1+x^2\right)}{\left(1+y^2\right)\left(1+xy\right)\left(1+x^2\right)}-\dfrac{x\left(x-y\right)\left(1+y^2\right)}{\left(1+x^2\right)\left(1+xy\right)\left(1+y^2\right)}$

$=\dfrac{\left(x-y\right)\left(y\left(1+x^2\right)-x\left(1+y^2\right)\right)}{\left(1+y^2\right)\left(1+xy\right)\left(1+x^2\right)}$

$=\dfrac{\left(x-y\right)\left(y+x^2y-x-xy^2\right)}{\left(1+y^2\right)\left(1+xy\right)\left(1+x^2\right)}$

$=\dfrac{\left(x-y\right)\left(xy\left(x-y\right)-\left(x-y\right)\right)}{\left(1+y^2\right)\left(1+xy\right)\left(1+x^2\right)}$

$=\dfrac{\left(x-y\right)^2\left(xy-1\right)}{\left(1+y^2\right)\left(1+xy\right)\left(1+x^2\right)}$

Vì $x,y\ge 1\to xy\ge 1\to xy-1\ge 0$

$\to \dfrac{\left(x-y\right)^2\left(xy-1\right)}{\left(1+y^2\right)\left(1+xy\right)\left(1+x^2\right)}\ge 0$

$\to \dfrac{1}{1+x^2}+\dfrac1{1+y^2}-\dfrac{2}{1+xy}\ge 0$

$\to \dfrac{1}{1+x^2}+\dfrac1{1+y^2}\ge\dfrac{2}{1+xy}$

$\to đpcm$