5. Given three triangles $\triangle, \triangle_{1}, \triangle_{2}$ with perimeters $p, p_{1}, p_{2}$ respectively. If $\triangle \backsim \triangle_{1} \backsim \triangle_{2}$, and the smaller two triangles $\triangle_{1}$ and $\triangle_{2}$ can be placed inside the larger triangle $\triangle$ without overlapping. Then the maximum value of $p_{1}+p_{2}$ is ( . . ).
(A) $p \cdot(\mathrm{B}) \sqrt{2} p$
(C) $\sqrt{3} p$
(D) $2 p$
5.13.

Let the perimeter of $\triangle$ be $p$, and the area be $S$; the perimeter of $\Delta$ be $p_{1}$, and the area be $S_{1}$; the perimeter of $\triangle_{2}$ be $p_{2}$, and the area be $S_{2}$. From the condition, we have $S \geqslant S_{1} + S_{2}$.
$$
\begin{array}{l}
\because \triangle \backsim \triangle_{1} \backsim \triangle_{2}, \\
\therefore \frac{S}{p^{2}}=\frac{S_{1}}{p_{1}^{2}}=\frac{S_{2}}{p_{2}^{2}}(=k) .
\end{array}
$$

Thus, $p^{2} \geqslant p_{1}^{2} + p_{2}^{2}$.
Since $p, p_{1}, p_{2} > 0$, we have..
$$
p^{2} \geqslant p_{1}^{2} + p_{2}^{2} \geqslant 2 p_{1} p_{2} \text{.}
$$

Adding the above two inequalities, we get
$$
2 p^{2} \geqslant p_{1}^{2} + p_{2}^{2} + 2 p_{1} p_{2} = (p_{1} + p_{2})^{2},
$$

which implies $\sqrt{2} p \geqslant p_{1} + p_{2}$.