5. Let $a, b, c$ be non-negative real numbers, satisfying $a+b+c=8, ab+bc+ca=16$. Let $m=\min \{ab, bc, ca\}$, then the maximum value of $m$ is $\qquad$ .
Let's assume $c \geqslant b \geqslant a$, then $3 c \geqslant a+b+c=8 \Rightarrow c \geqslant \frac{8}{3}$,
$$
\left\{\begin{array}{l}
a+b=8-c, \\
a^{2}+b^{2}=32-c^{2}
\end{array} \Rightarrow 2\left(32-c^{2}\right) \geqslant(8-c)^{2} \Rightarrow 3 c^{2}-16 c \leqslant 0 \Rightarrow \frac{8}{3} \leqslant c \leqslant \frac{16}{3}\right. \text {. }
$$

Since $b c \geqslant a c \geqslant a b$, and $a b=16-c(a+b)=16-c(8-c)=(c-4)^{2} \leqslant \frac{16}{9}$. Therefore, the maximum value of $m$ is $\frac{16}{9}$, at which point $(a, b, c)=\left(\frac{4}{3}, \frac{4}{3}, \frac{16}{3}\right)$.