Question

Chứng minh rằng:

x^2+4y^2+z^2-2x-6z+8y+15>0(> ko phải là >=)

 

Answer 1

\(x^2+4y^2+z^2-2x-6z+8y+15=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)

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

thấy: \(\left(x-1\right)^2\ge0;\left(2y+2\right)^2\ge0;\left(z-3\right)^2\ge0\)

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

\(\Rightarrow\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\) (đpcm)