ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)
\(\Rightarrow x^2+y^2+z^2=9\)
áp dung bu nhi a \(2\left(y^2+z^2\right)\ge\left(y+z\right)^2\)
\(\Leftrightarrow2\left(9-x^2\right)\ge\left(5-x\right)^2\)
\(\Leftrightarrow18-2x^2\ge25-10x+x^2\)
\(\Leftrightarrow0< =3x^2-10x+7\)
suy ra 1<=x<=7/3