$(x^2-2)^2\geq 0\Leftrightarrow x^4+4\geq 4x^2$ So $5yz\geq 4x^2+y^2$.Similarly $5xz\geq 4y^2+z^2,5xy\geq 4z^2+x^2$ so $5(xy+yx+xz)\geq 5(x^2+y^2+z^2)$ This is true only for $x=y=z$ and we must have $(x^2-2)^2=0$ so $(x,y,z)=(\sqrt{2},\sqrt{2},\sqrt{2}),(-\sqrt{2},-\sqrt{2},-\sqrt{2})$