5. Notice that $\frac{x^{2}-x y+y^{2}}{x^{2}+x y+y^{2}} \geqslant \frac{1}{3} \Leftrightarrow 3\left(x^{2}-x y+y^{2}\right) \geqslant x^{2}+x y+y^{2} \Leftrightarrow 2(x-$ $y)^{2} \geqslant 0$. Then $\frac{x^{3}+y^{3}}{x^{2}+x y+y^{2}}=\frac{x^{2}-x y+y^{2}}{x^{2}+x y+y^{2}}(x+y) \geqslant \frac{x+y}{3}$. Therefore, $\frac{x^{3}+y^{3}}{x^{2}+x y+y^{2}}+$
$$\begin{array}{l} \frac{y^{3}+z^{3}}{y^{2}+y z+z^{2}}+\frac{z^{3}+x^{3}}{z^{2}+z x+x^{2}} \geqslant \frac{1}{3}(x+y)+\frac{1}{3}(y+z)+\frac{1}{3}(z+x)=\frac{2}{3}(x+ \\ y+z) \geqslant 2 \sqrt[3]{x y z}=2 \end{array}$$