Question

Chứng minh:

\(x^2+y^2+z^2+14\ge4x-2y-6z\)

Answer 1

Ta có:

\(x^2+y^2+z^2-4x+2y+6z\)

\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+\) \(\left(z^2+6z+9\right)\)

\(=\left(x-2\right)^2+\left(y+1\right)^2+\left(z+3\right)^2\)

Mà : \(\left(x-2\right)^2\ge0\forall x\)

        \(\left(y+1\right)^2\ge0\forall y\)

          \(\left(z+3\right)^2\ge0\forall z\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z+3\right)^2\ge0\forall x;y;z\) ( luôn đúng )

\(\Rightarrow x^2+y^2+z^2+14\ge4x-2y-6z\left(đpcm\right)\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x-2=0\\y+1=0\\z+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\\z=-3\end{cases}}\)

Vậy ....