Question

Cho \(\hept{\begin{cases}x,y,z\in\left|0;2\right|\\x+y+z=3\end{cases}}\)

CMR :\(x^2+y^2+z^2\le5\)

Answer 1

ko bít ^_^!

!@#$%^&*()_+L:"><?/.,~`

= ???????????/

Answer 2

\(0\le x,y,z\le2\Leftrightarrow xyz+\left(2-x\right)\left(2-y\right)\left(2-z\right)\ge0\)

\(\Leftrightarrow8-4\left(x+y+z\right)+2\left(xy+yz+zx\right)\ge0\)

\(\Leftrightarrow xy+yz+xz\ge2\)

xét \(x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+xz\right)=9-2\left(xy+yz+xz\right)\le9-2.2=5\)

Dấu = xảy ra khi \(\left(x;y;z\right)=\left(0;1;2\right)\)và các hoán vị