(for a right-angle triangle the problem is obvios) $ [A'B'C'] = \frac{R^2}{4} \left ( \tan A + \tan B + \tan C \right ) = \frac{R^2}{4} \tan A \tan B \tan C$ and $ [ABC]= 2R^2 \sin A \sin B \sin C$. Then we have to prove that $ \tan A \tan B \tan C \ge 8 \sin A \sin B \sin C$ that is equal to the well-know $ \cos A \cos C \cos C \le \frac{1}{8}$. We have the equality iff $ A = B = C$.