31. Let $x, y, z$ be real numbers such that
$$
x+y+z=1 \quad \text { and } \quad x^{2}+y^{2}+z^{2}=1
$$

Let $m$ denote the minimum value of $x^{3}+y^{3}+z^{3}$. Find $9 m$.
31. Answer: 5 .
It is clear that $|x|,|y|,|z| \leq 1$. Note that
$$
0=x+y+z-x^{2}-y^{2}-z^{2}=x(1-x)+y(1-y)+z(1-z) .
$$

Without loss of generality, assume that $z \leq 0$. Then $x=(1-y)+(-z) \geq 0$ and similarly $y \geq 0$. Since
$$
1-z^{2}=x^{2}+y^{2} \geq \frac{(x+y)^{2}}{2}=\frac{(1-z)^{2}}{2},
$$
we have $-1 / 3 \leq z \leq 0$. On the other hand,
$$
\begin{aligned}
x^{3}+y^{3}+z^{3} & =(x+y)\left[\frac{3\left(x^{2}+y^{2}\right)-(x+y)^{2}}{2}\right]+z^{3} \\
& =(1-z)\left[\frac{3\left(1-z^{2}\right)-(1-z)^{2}}{2}\right]+z^{3}=1-3 z^{2}+3 z^{3}
\end{aligned}
$$
increases as $z$ increases. The minimum $m=5 / 9$ is obtained at $z=-1 / 3$ and $x=y=2 / 3$.