\(\text{Sử dụng AM-GM, ta có}\)
\(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
\(xy+yz+xz\le x^2+y^2+z^2\)
\(\text{Cộng theo vế, ta được}\)
\(6=x+y+z+xy+yz+xz\le\sqrt{3\left(x^2+y^2+z^2\right)+x^2+y^2+z^2}\)
Suy ra\(x^2+y^2+z^2\ge3\)
\(x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z\)
\(\Rightarrow x^2+y^2+z^2+3\ge2x+2y+2z\Rightarrow\frac{x^2+y^2+z^2}{2}+\frac{3}{2}\ge x+y+z\)
\(x^2+y^2\ge2xy;y^2+z^2\ge2yz;z^2+x^2\ge2zx\)
\(\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\)
Khi đó:\(\frac{3}{2}\left(x^2+y^2+z^2\right)+\frac{3}{2}\ge x+y+z+xy+yz+zx=6\)
\(\Rightarrow x^2+y^2+z^2+1\ge4\Rightarrow x^2+y^2+z^2\ge3\)
\(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=\frac{1}{3}\)
Dấu "=" xảy ra tại x=y=z=1/3
dấu "=" bài trước là x=y=z=1 nhé !
