Đáp án: $ A\ge 6$
Giải thích các bước giải:
Ta có:
$A=\dfrac{3+x^2}{y+z}+\dfrac{3+y^2}{z+x}+\dfrac{3+z^2}{x+y}$
$\to A=3(\dfrac{1}{x+y}+\dfrac1{y+z}+\dfrac1{z+x})+(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y})$
$\to A=3(\dfrac{1^2}{x+y}+\dfrac{1^2}{y+z}+\dfrac{1^2}{z+x})+(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y})$
$\to A\ge 3\cdot \dfrac{(1+1+1)^2}{(x+y)+(y+z)+(z+x)}+\dfrac{(x+y+z)^2}{(y+z)+(z+x)+(x+y)}$
$\to A\ge 3\cdot \dfrac{9}{2(x+y+z)}+\dfrac{(x+y+z)^2}{2(x+y+z)}$
$\to A\ge 6$
Dấu = xảy ra khi $x=y=z=1$