Question

[Trần Khánh Hưng]

chứng minh x^2 +y^2+z^2+2x-4y-6z+14 >= 0 với mọi x y z

Answer 1

[Chu Công Đức]

\(x^2+y^2+z^2+2x-4y-6z+14\)

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

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

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

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

hay \(x^2+y^2+z^2+2x-4y-6z+14\ge0\)\(\forall x,y,z\)