Proof: As shown in Figure 1, the plane region of acute triangle $\triangle ABC$ can be divided into three parts, namely $\angle AOB$, $\angle BOC$, and $\angle COA$. Suppose the center $D$ is within the region $\angle BOC$, and connect $AD$, $BD$, and $CD$. Clearly,
$$\angle ABD \geq \angle ABO, \quad \angle ACD \geq \angle ACO,$$
and among $\angle BAD$ and $\angle CAD$, at least one is not less than $\angle BAO$ or $\angle CAO$.

Assume $\angle BAD \geq \angle BAO$, then point $O$ must be inside $\triangle ABD$. Clearly, $R \geq BD$ and $R \geq AD$. Therefore,
$$2R \geq BD + AD \geq OA + OB = 2r.$$

Thus, $R \geq r$.

It is well known that the diameter of the circumcircle of a right triangle is its hypotenuse, so the smallest circle that can cover a right triangle is its circumcircle.

For an obtuse triangle, is its circumcircle also the smallest circle that can cover the triangle? The answer is no. The smallest circle that can cover an obtuse triangle is the circle with the side opposite the obtuse angle as its diameter.