5. A sphere with radius $R$ contains 4 smaller spheres with radius $r$. The maximum value of the radius $r$ of the smaller spheres is $\qquad$ . (Express the answer in terms of $R$)
Given a regular tetrahedron $ABCD$ with edge length $2r$, the radius of the circumscribed sphere of $ABCD$ is $\frac{\sqrt{6} r}{2}$. Therefore, $\frac{\sqrt{6} r}{2} + r \leqslant R \Rightarrow (\sqrt{6} + 2) r \leqslant 2R$ $\Rightarrow r \leqslant \frac{2R}{\sqrt{6} + 2} = (\sqrt{6} - 2)R$.
Thus, the maximum value of the radius $r$ of the smaller sphere is $(\sqrt{6} - 2)R$.