Áp dụng BĐT \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\Leftrightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
\(\Rightarrow Q\le\sqrt{3\left(x+y+y+z+z+x\right)}=\sqrt{6\left(x+y+z\right)}=\sqrt{6}\)
\(Q_{max}=\sqrt{6}\) khi \(x=y=z=\frac{1}{3}\)