* 4. Real numbers $a, b, c$ satisfy $a+b-2 c=-3, a^{2}+b^{2}+2 c^{2}$ $+8 c=5$. Then the minimum value of $a b$ is $\qquad$.
4. $\frac{25}{16}$. From the given conditions, we have $a+b=2 c-3, a b=c^{2}-2 c+2$.

Therefore, $a$ and $b$ are the real roots of the equation
$$
t^{2}-(2 c-3) t+\left(c^{2}-2 c+2\right)=0
$$

Thus,
$$
\begin{array}{c}
\Delta=(2 c-3)^{2}-4\left(c^{2}-2 c+2\right) \geqslant 0, \\
c \leqslant \frac{1}{4} .
\end{array}
$$

Also, $a b=f(c)=c^{2}-2 c+2$ is a decreasing function when $c<1$, so when $c=\frac{1}{4}$, $a b$ has the minimum value $f\left(\frac{1}{4}\right)=\frac{25}{16}$.