Không mất tính tổng quát, giả sử \(x\le y\le z\)
\(\Rightarrow\frac{3}{2}=x+y+z\le3z\Rightarrow\frac{1}{2}\le z\le1\)
\(P=x^2+y^2+z^2\le x^2+2xy+y^2+z^2=\left(x+y\right)^2+z^2\)
\(\Rightarrow P\le\left(\frac{3}{2}-z\right)^2+z^2=2z^2-3z+\frac{9}{4}=\left(z-1\right)\left(2z-1\right)+\frac{5}{4}\)
Do \(\frac{1}{2}\le z\le1\Rightarrow\left\{{}\begin{matrix}z-1\le0\\2z-1\ge0\end{matrix}\right.\) \(\Rightarrow\left(z-1\right)\left(2z-1\right)\le0\)
\(\Rightarrow P\le\frac{5}{4}\Rightarrow P_{max}=\frac{5}{4}\) khi \(\left(x;y;z\right)=\left(0;\frac{1}{2};1\right)\) và các hoán vị