Question

X^2+y^2+z^2+2x+2y-2z+5 > 0 

Cm ho minh vs mn

Answer 1

\(x^2+y^2+z^2+2x+2y-2z+5\)

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

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

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

\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z-1\right)^2\ge2\forall x;y;z\)

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

Answer 2

Sửa hộ dòng thứ 5 là \(\left(x+1\right)^2+\left(y+1\right)^2+\left(z-1\right)^2+2\ge2\forall x;y;z\)nha