Đơn giản là C-S:
\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{y+z+x+z+x+y}=\frac{x+y+z}{2}\)
Hoặc làm theo AM-GM:
\(\frac{x^2}{y+z}+\frac{y+z}{4}\ge x\) ; \(\frac{y^2}{x+z}+\frac{x+z}{4}\ge y\); \(\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\)
Cộng vê với vế:
\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}+\frac{1}{2}\left(x+y+z\right)\ge x+y+z\)
\(\Leftrightarrow\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{1}{2}\left(x+y+z\right)\)
Dấu "=" xảy ra khi \(x=y=z\)
