tìm GTNN

E

eye_smile

Có: $xy+yz+zx=\dfrac{(x+y+z)^2-1}{2}$

$P=x+y+z+\dfrac{(x+y+z)^2-1}{2}$

Đặt $x+y+z=A$ \Rightarrow $P=A+\dfrac{A^2-1}{2}$

\Leftrightarrow $2P=2A+A^2-1=(A+1)^2-2 \ge -2$

\Leftrightarrow $P \ge -1$

Dấu "=" xảy ra \Leftrightarrow $x+y+z=-1;x^2+y^2+z^2=1$
 
You must log in or register to reply here.
Top Bottom