`~rai~`
\(\text{Áp dụng BĐT Bunhiacopxki,ta có:}\\(x^2+y^2+z^2)(1^2+1^2+1^2)\ge (x.1+y.1+z.1)^2\\\Leftrightarrow 3(x^2+y^2+z^2)\ge (x+y+z)^2\\\Leftrightarrow 3(x^2+y^2+z^2)\ge 1^2\\\Leftrightarrow x^2+y^2+z^2\ge \dfrac{1}{3}\\\text{Dấu "=" xảy ra}\Leftrightarrow \dfrac{x}{1}=\dfrac{y}{1}=\dfrac{z}{1}\\\Leftrightarrow x=y=z=\dfrac{1}{3}.\)