11. If positive real numbers $x, y, z$ satisfy $x^{3}+y^{3}+z^{3}-3 x y z=1$, find the minimum value of $x^{2}+y^{2}+z^{2}$.
\begin{array}{l}\text { 11. Solution: Let } x+y+z=k>0, \because x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)=1 \text {. } \\ \therefore x^{2}+y^{2}+z^{2}-x y-y z-z x=\frac{1}{k} . \because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x=k^{2} \text {. } \\ \therefore x y+y z+z x=\frac{k^{3}-1}{3 k} . \therefore x^{2}+y^{2}+z^{2}=\frac{k^{3}+2}{3 k}=\frac{k^{2}}{3}+\frac{1}{3 k}+\frac{1}{3 k} \geqslant 3 \sqrt[3]{\frac{k^{2}}{3} \times \frac{1}{3 k} \times \frac{1}{3 k}}=1 \text {. } \\ \therefore\left(x^{2}+y^{2}+z^{2}\right)_{\min }=1 \text { (when } x=y=0, z=1 \text { the equality holds). } \\\end{array}