Môn Toán Lớp 9: cho x,y,z>0 .CMR (x^2/(y+z))+(y^2/(x+z))+(z^2/(x+y)) >= (x+y+z)/2

Áp dụng BĐT Svacxơ (BĐT cộng mẫu), ta có:

$\dfrac{x^2}{y + z}$ + $\dfrac{y^2}{x + z}$ + $\dfrac{z^2}{x + y}$ ≥ $\dfrac{(x + y + z)^2}{y + z + x + z + x + y}$ = $\dfrac{(x + y + z)^2}{2(x + y + z)}$ = $\dfrac{x + y + z}{2}$

Dấu “=” xảy ra khi: x = y = z