Question

Cho x,y,z là những số thực thỏa mãn \(\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2=1\)

CMR \(|2x-3y+4z-20|\le\sqrt{29}\)

Answer 1

\(\left[2\left(x-1\right)-3\left(y+2\right)+4\left(z-3\right)\right]^2\le\left(2^2+3^2+4^2\right)\left[\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2\right]\)

\(\Rightarrow\left(2x-3y+4z-20\right)^2\le29\)

\(\Rightarrow\left|2x-3y+4z-20\right|\le\sqrt{29}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2=1\\\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{4}\end{matrix}\right.\)