Question

Cho x,y,z là 3 số dương thỏa mãn xy + yz + zx - xyz = 0, Tìm gtnn:

A= \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)

Answer 1

\(xy+yz+zx-xyz=0\Leftrightarrow xy+yz+zx=xyz\Leftrightarrow;\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1;\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\Rightarrow1\ge\frac{9}{x+y+z}\Rightarrow x+y+z\ge9\)

\(A=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{\left(x+y+z\right)}{2}\ge\frac{9}{2}\)

Dấu "=" xayr ra khi x=y=z=3