5. For any $\square ABC$, with the incenter $I$. Then, when $AB + AC \geqslant 2BC$, the relationship between the circumradius $R_{1}$ of $\square ABC$ and the circumradius $R_{2}$ of $\square IBC$ is ().
$$
\begin{array}{ll}
\text { ( A ) } R_{1} \geqslant R_{2} & \text { ( B ) } R_{1}>R_{2} \\
\text { ( C) } R_{1} \leqslant \sqrt{3} R_{2} & \text { ( D) } R_{2} \leqslant R_{1} \leqslant 2 R_{2}
\end{array}
$$
5.(A).

As shown in Figure 4, extend $A I$ to $D$ such that $D$ lies on the circumcircle of $\triangle ABC$. Connect $C D$ and $B D$. It can be proven through angle transformation that
$$
D I=D B=D C .
$$

Thus, $R_{2}=D C$.
Let $A D$ and $B C$ intersect at $M$.
From $\angle 1=\angle 2$ and $\angle A B C=\angle A D C$, we get
$\triangle A B M \sim \triangle A D C$.
By the Angle Bisector Theorem, we have
$$
\frac{A B}{B M}=\frac{A C}{C M}=\frac{A B+A C}{B M+C M}=\frac{A B+A C}{B C} \geqslant 2 .
$$

Thus, $\frac{A D}{D C}=\frac{A B}{B M} \geqslant 2$,
$$
2 R_{1} \geqslant A D \geqslant 2 D C=2 R_{2} \text {, }
$$

Therefore, $R_{1} \geqslant R_{2}$.
Moreover, when $\triangle A B C$ is an equilateral triangle, we have $R_{1}=R_{2}$.
When $\frac{A C+A B}{B C}$ approaches $+\infty$, $2 R_{1}$ will be much greater than $2 R_{2}$.
Hence, it is impossible for $R_{1} \leqslant 2 R_{2}$.