D
dinhnam9f
Cho x,y,z > 0 và x + y + z = 1. Tìm min :
[TEX]P = \frac{x^2(y+z)}{yz} + \frac{y^2(z+x)}{zx} + \frac{z^2(x+y)}{xy} [/TEX]
Using
[TEX] \frac{1}{a}+\frac{1}{b} \ge \frac{4}{a+b}[/TEX]
or
[TEX] LHS \ge 4\sum \frac{x^2}{y+z} \ge \frac{4(x+y+z)^2}{2(x+y+z)}=2(a+b+c)=2[/TEX]
We have done !!!