5. Let the function $f(x)=a+\sqrt{-x^{2}-4 x}, g(x)$ $=\frac{4}{3} x+1$. If $f(x) \leqslant g(x)$ always holds, then the range of the real number $a$ is $\qquad$.
5. $a \leqslant-5$.

From the problem, we have
$$
\begin{array}{l}
\sqrt{-x^{2}-4 x} \leqslant \frac{4}{3} x+1-a . \\
\text { Let } y_{1}=\sqrt{-x^{2}-4 x}, \\
y_{2}=\frac{4}{3} x+1-a .
\end{array}
$$

Then equation (2) represents the upper half of the circle $(x+2)^{2}+y_{1}^{2}=4$ (including the endpoints), and equation (3) represents a line with a slope of $\frac{4}{3}$. For equation (1) to always hold, the circular arc represented by equation (2) must be below the line (3).

When the line is tangent to the circle, the distance from the center of the circle to the line equals the radius, thus, we have $\frac{|-8+3-3 a|}{\sqrt{4^{2}+(-3)^{2}}}=2$, which gives $a=-5$.

Therefore, for $f(x) \leqslant g(x)$ to always hold, the range of real number $a$ is $a \leqslant-5$.