5. Let $a, b, c$ be non-negative real numbers, satisfying
$$
a+b+c=8, \quad ab+bc+ca=16 \text{. }
$$

Let $m=\min \{ab, bc, ca\}$. Then the maximum possible value of $m$ is . $\qquad$
5. $\frac{16}{9}$.

Assume $c \geqslant b \geqslant a$. Then
$$
3 c \geqslant a+b+c=8 \Rightarrow c \geqslant \frac{8}{3} \text {. }
$$
$$
\begin{array}{l}
\text { By } 0 \leqslant(a-b)^{2}=(a+b)^{2}-4 a b \\
=(8-c)^{2}-4(16-c(a+b)) \\
=(8-c)^{2}-4(16-c(8-c)) \\
=c(16-3 c) \\
\Rightarrow 0 \leqslant c \leqslant \frac{16}{3} .
\end{array}
$$

By $b c \geqslant c a \geqslant a b$, hence
$$
m=\min \{a b, b c, c a\}=a b .
$$

And $a b=16-c(a+b)=(c-4)^{2}$, combining $\frac{8}{3} \leqslant c \leqslant \frac{16}{3}$, we know $a b \leqslant \frac{16}{9}$.
When $(a, b, c)=\left(\frac{4}{3}, \frac{4}{3}, \frac{16}{3}\right)$, $m$ reaches its maximum value.