5. In $\triangle A B C$, $A B=3, A C=4, B C=5, I$ is the incenter of $\triangle A B C$, and $P$ is a point within $\triangle I B C$ (including the boundary). If $\overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}(\lambda ; \mu \in \mathbf{R})$, then the minimum value of $\lambda+\mu$ is
5. $\frac{7}{12}$.

Let $A P$ intersect $B C$ at point $Q$. Then $\overrightarrow{A Q} \triangleq t \overrightarrow{A P}=t \lambda \overrightarrow{A B}+t_{\mu} \overrightarrow{A C}$.
Since $Q, B, C$ are collinear,
$$
\Rightarrow t \lambda+t \mu=1 \Rightarrow \lambda+\mu=\frac{1}{t} \text {. }
$$

When point $P$ coincides with $I$, $t$ is maximized, at this time,
$$
\frac{A P}{P Q}=\frac{7}{5} \Rightarrow \frac{A Q}{A P}=\frac{12}{7} \text {. }
$$

Therefore, the maximum value of $\lambda+\mu=\frac{1}{t}$ is $\frac{7}{12}$.