cho 0< x,y,z \leq 1 x+ y \geq 1 + z Tìm Min P

V

vuive_yeudoi

Có
$$ \left( x+y \right) - \left( xy+z^2 \right)=\left( x+y-1-z \right)+z \cdot \left( 1-z \right) + \left( 1-xy \right) \ge 0 $$
Vậy
$$ P \ge \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \ge \frac{3}{2} $$
Tại $ \displaystyle \left( x,y,z \right)= \left( 1,1,1 \right) $ thì
$$ P = \frac{3}{2} $$
Vậy nên
$$ \min P =\frac{3}{2} $$
 
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