5. The solution set of the inequality $\sqrt{x^{2}+4 x} \leqslant 4-\sqrt{16-x^{2}}$ is
5. $\{-4,0\}$.
$\left\{\begin{array}{l}x^{2}+4 x \geqslant 0 \\ 16-x^{2} \geqslant 0,\end{array}\right.$ we get $x=-4$ or $x \in[0,4]$. Clearly, $x=-4$ is a solution to the inequality. When $x \in[0,4]$, $4 \geqslant \sqrt{x^{2}+4 x}+\sqrt{16-x^{2}}=\sqrt{x+4}(\sqrt{x}+\sqrt{4-x}) \geqslant 2(\sqrt{x}+\sqrt{4-x})=$ $2 \sqrt{(\sqrt{x}+\sqrt{4-x})^{2}}=2 \sqrt{4+2 \sqrt{x(4-x)}} \geqslant 2 \sqrt{4}=4$, equality holds when $x=0$.