8. In a regular tetrahedron \(ABCD\) with edge length 1, two tangent spheres are placed inside, one of which is tangent to the three faces passing through point \(A\), and the other is tangent to the three faces passing through point \(B\). Then the minimum value of the sum of their radii is \(\qquad\)
8. $\frac{\sqrt{6}-1}{5}$.

Let the centers of the two spheres be $P, Q$, and their radii be $x, y$, respectively. Without loss of generality, assume that sphere $P$ is tangent to the three faces passing through point $A$, and sphere $Q$ is tangent to the three faces passing through point $B$, then $A P=3 x, B Q=3 y$.
Let the projections of $P, Q$ on the plane $A B C$ be $M, N$, respectively, then
$$
M N=\sqrt{(x+y)^{2}-(x-y)^{2}}=2 \sqrt{x y} .
$$

Let the projections of $M, N$ on the line $A B$ be $U, V$, respectively, then $A M=2 \sqrt{2} x, M U=\sqrt{2} x$, $A U=\sqrt{6} x, B N=2 \sqrt{2} y, N V=\sqrt{2} y, B V=\sqrt{6} y$. In the right trapezoid $M N U V$, we have $M N^{2}=$ $(M U-N V)^{2}+U V^{2}$, so
$$
4 x y=(\sqrt{2} x-\sqrt{2} y)^{2}+(1-\sqrt{6} x-\sqrt{6} y)^{2},
$$

Simplifying, we get
$$
12 x y=8(x+y)^{2}-2 \sqrt{6}(x+y)+1 .
$$

By the AM-GM inequality, we have $12 x y \leqslant 3(x+y)^{2}$, substituting into the above equation, we solve to get $x+y \geqslant \frac{\sqrt{6}-1}{5}$. The equality can clearly be achieved. Therefore, the minimum value sought is $\frac{\sqrt{6}-1}{5}$.