theo định lí đi dép tổ ong thì 2 trong 3 số x-2;y-2;z-2 cùng dấu
giả sử \(\left(x-2\right)\left(y-2\right)\ge0\Leftrightarrow xy-2\left(x+y\right)+4\ge0\)
\(\Leftrightarrow xy-2\left(6-z\right)+4\ge0\)
<=>xy-8+2z>(=)0
<=>xyz+2z^2-8z>(=)0
<=>xyz>(=)8z-2z^2
\(x^2-xy+y^2\ge\frac{x^2+y^2}{2}\ge\frac{\left(x+y\right)^2}{4}=\frac{\left(6-z\right)^2}{4}=\frac{z^2}{4}-3z+9\)
xz+yz=z(x+y)=x(6-z)=6z-z2
\(\Rightarrow x^2+y^2+z^2-xy-yz-zx+xyz\ge\frac{z^2}{4}-3z+9+z^2+z^2-6z+8z-z^2=\frac{z^2}{4}-z+9=\left(\frac{z}{2}-1\right)^2+8\ge8\)
