Question

Cho x,y,z thuộc đoạn [1;2] thỏa mãn x²+y²+z²=6.CMR x+y+z\(\ge0\)

Answer 1

\(1\le x;y;z\le2\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x-2\right)\le0\\\left(y-1\right)\left(y-2\right)\le0\\\left(z-1\right)\left(z-2\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{x^2+2}{3}\\y\ge\frac{y^2+2}{3}\\z\ge\frac{z^2+2}{3}\end{matrix}\right.\) \(\Rightarrow x+y+z\ge\frac{x^2+y^2+z^2+6}{3}=4\)

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