5. Given real numbers $a, b, c$ satisfy $a+b+c=0$, $a^{2}+b^{2}+c^{2}=1$. Then the maximum value of $a b c$ is $\qquad$
5. $\frac{\sqrt{6}}{18}$.

Solution 1: From $a^{2}+b^{2}+c^{2}=1$, we know that $a^{2}, b^{2}, c^{2}$ must have one that is no greater than $\frac{2}{3}$. Without loss of generality, let $c^{2} \leqslant \frac{2}{3}$. Let $c=\sqrt{\frac{2}{3}} \cos \theta$, then
$$
\begin{array}{l}
a b c=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] c \\
=\frac{1}{2}\left[c^{2}-\left(1-c^{2}\right)\right] c=\frac{c}{2}\left(2 c^{2}-1\right) \\
=\frac{1}{2} \times \sqrt{\frac{2}{3}} \cos \theta\left(2 \times \frac{2}{3} \cos ^{2} \theta-1\right) \\
=\frac{\sqrt{6}}{18}\left(4 \cos ^{3} \theta-3 \cos \theta\right) \\
=\frac{\sqrt{6}}{18} \cos 3 \theta \leqslant \frac{\sqrt{6}}{18} .
\end{array}
$$

When $\cos 3 \theta=1$, take $\theta=0$, then
$$
c=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3} \text {. }
$$

Substituting into the known conditions, we get
$$
\left\{\begin{array}{l}
a+b=-\frac{\sqrt{6}}{3}, \\
a^{2}+b^{2}=1-\frac{2}{3}=\frac{1}{3} .
\end{array}\right.
$$

We can solve for $a=b=-\frac{\sqrt{6}}{6}$.
Therefore, when $a=b=-\frac{\sqrt{6}}{6}, c=\frac{\sqrt{6}}{3}$, $a b c$ achieves the maximum value $\frac{\sqrt{6}}{18}$.
Solution 2: Clearly, $a, b, c$ are not all $0$, and there must be two of $a, b, c$ whose product is non-negative. Without loss of generality, let $a b \geqslant 0$, then
$$
\begin{array}{l}
1=a^{2}+b^{2}+c^{2} \\
\geqslant \frac{(a+b)^{2}}{2}+c^{2}=\frac{c^{2}}{2}+c^{2}=\frac{3}{2} c^{2} .
\end{array}
$$

Thus, $|c| \leqslant \sqrt{\frac{2}{3}}$.
$$
\begin{array}{l}
\text { Also, }|a b c|=a b|c| \\
\leqslant\left(\frac{a+b}{2}\right)^{2}|c| \text { (using } a b \geqslant 0 \text { ) } \\
=\frac{c^{2}}{4}|c|=\frac{1}{4}|c|^{3} \\
\leqslant \frac{1}{4} \times \frac{2}{3} \sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{18},
\end{array}
$$

The equality holds if and only if $a=b$ and $|c|=\sqrt{\frac{2}{3}}$.
Taking $c=\sqrt{\frac{2}{3}}$, substituting into the known conditions, we get
$$
\left\{\begin{array}{l}
2 a+\sqrt{\frac{2}{3}}=0, \\
2 a^{2}+\frac{2}{3}=1 .
\end{array}\right.
$$

Solving, we get $a=b=-\frac{\sqrt{6}}{6}$.
Thus, when $a=b=-\frac{\sqrt{6}}{6}, c=\frac{\sqrt{6}}{3}$, $a b c$ achieves the maximum value $\frac{\sqrt{6}}{18}$.