Giải thích các bước giải:
Ta có :
$A=\dfrac{1+x^2}{1+y+z^2}+\dfrac{1+y^2}{1+z+x^2}+\dfrac{1+z^2}{1+x+y^2}$
$\to A=\dfrac{1+z+x^2}{1+y+z^2}+\dfrac{1+x+y^2}{1+z+x^2}+\dfrac{1+y+z^2}{1+x+y^2}-(\dfrac{z}{1+y+z^2}+\dfrac{x}{1+z+x^2}+\dfrac{y}{1+x+y^2})$
$\to A\ge3\sqrt[3]{\dfrac{1+z+x^2}{1+y+z^2}.\dfrac{1+x+y^2}{1+z+x^2}.\dfrac{1+y+z^2}{1+x+y^2}}-(\dfrac{z}{1+y+z^2}+\dfrac{x}{1+z+x^2}+\dfrac{y}{1+x+y^2})$
$\to A\ge3-(\dfrac{z}{1+y+z^2}+\dfrac{x}{1+z+x^2}+\dfrac{y}{1+x+y^2})$
$\to A\ge3-(\dfrac{z}{y+2z}+\dfrac{x}{z+2x}+\dfrac{y}{x+2y})$
$\to A\ge3-(\dfrac{1}{2}-\dfrac{y}{2(y+2z)}+\dfrac{1}{2}-\dfrac{z}{2(z+2x)}+\dfrac{1}{2}-\dfrac{x}{2(x+2y)})$
$\to A\ge3-\dfrac{3}{2}+\dfrac{1}{2}(\dfrac{y}{y+2z}+\dfrac{z}{z+2x}+\dfrac{x}{x+2y})$
$\to A\ge\dfrac{3}{2}+\dfrac{1}{2}(\dfrac{y^2}{y^2+2yz}+\dfrac{z^2}{z^2+2xz}+\dfrac{x^2}{x^2+2xy})$
$\to A\ge\dfrac{3}{2}+\dfrac{1}{2}(\dfrac{(x+y+z)^2}{y^2+2yz+z^2+2xz+x^2+2xy})$
$\to A\ge\dfrac{3}{2}+\dfrac{1}{2}(\dfrac{(x+y+z)^2}{(x+y+z)^2})$
$\to A\ge 2$
Dấu = xảy ra khi $x=y=z=1$
