3. (Canada 1)

Proof: Let $R, r, s, \triangle$ denote the circumradius, inradius, semiperimeter, and area of $\triangle ABC$, respectively, and let $\triangle^{\prime}$ denote the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$. Then,
$$\begin{array}{l} \triangle=\frac{a b c}{4 R}=2 R^{2} \sin A \sin B \sin C, \\ \triangle^{\prime}=2 R^{2} \sin A^{\prime} \sin B^{\prime} \sin C^{\prime} . \end{array}$$
The angle at $C^{\prime}$ in $\triangle A^{\prime} B^{\prime} C^{\prime}$ is $\frac{A+B}{2}=90^{\circ}-\frac{C}{2}$. Substituting this into (2) yields
$$\Delta^{\prime}=2 R^{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} .$$

Thus,

For $\sin \frac{B}{2}$ and $\sin \frac{C}{2}$, there are similar formulas. Substituting them into (4) gives
$$\begin{array}{l} - \sqrt{\frac{(s-c)(s-a)}{a c}} \cdot \sqrt{\frac{(s-a)(s-b)}{a b}} \\ =8 \frac{(s-a)}{a b c} . \\ \end{array}$$

By Heron's formula, $\triangle=\sqrt{s(s-a)(s-b)(s-c)}$, so (5) becomes
$$\frac{\triangle}{\Delta^{\prime}}=8 \frac{\triangle^{2}}{s a b c} \text {. }$$

Since $\triangle / s=r$ and $\triangle / a b c=1 / 4 R$, substituting these into (6) gives
$$\frac{\Delta}{\triangle^{\prime}}=\frac{2 r}{R} \text {. }$$

By Euler's inequality, $2 r / R \leqslant 1$, which completes the proof.