6 Inside a sphere of radius $R$, there are 4 smaller spheres of the same radius $r$. The possible maximum value of the radius $r$ of the smaller spheres is ( ).
(A) $\frac{\sqrt{3}}{2+\sqrt{3}} R$
(B) $\frac{\sqrt{6}}{3+\sqrt{6}} R$
(C) $\frac{1}{1+\sqrt{3}} R$
(D) $\frac{\sqrt{5}}{2+\sqrt{5}} R$
(6) B Prompt: As shown in the figure, connect the centers of the four small balls $\mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}, \mathrm{O}_{4}$ to form a tetrahedron. Let the projection of $\mathrm{O}_{1}$ on the base $\mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ be $\mathrm{H}$. The center of the large ball $O$ is clearly on $O_{1} \mathrm{H}$. Connect $\mathrm{O}_{2} \mathrm{H}, \mathrm{O}_{3} \mathrm{H}, \mathrm{OO}_{3}$ and set $O_{1} O=x$. $O_{2} H=\frac{\sqrt{3}}{3} \cdot 2 r$,
$$
\begin{array}{c}
O_{1} H=\sqrt{(2 r)^{2}-\left(\frac{2 \sqrt{3}}{3} r\right)^{2}}=\frac{2}{3} \sqrt{6} r, \\
O_{3} H=O_{2} H=\frac{2 \sqrt{3}}{3} r .
\end{array}
$$

In the right triangle $\triangle O O_{3} \mathrm{H}$, $O H=\frac{2}{3} \sqrt{6} r-x, O O_{3}=O O_{1}=x$. By the Pythagorean theorem,
$$
x^{2}=\left(\frac{2}{3} \sqrt{6} r-x\right)^{2}+\left(\frac{2}{3} \sqrt{3} r\right)^{2},
$$

Solving for $x$ gives $x=\frac{\sqrt{6}}{2} r$. Since $x+r=R, \frac{\sqrt{6}}{2} r+r=R$, we have $r=\frac{\sqrt{6}}{3+\sqrt{6}} R$.