Problem 1496

Let $a=\sqrt{2 \tan A}, b=\sqrt{2 \tan B}, c=\sqrt{2 \tan C}$, where $A, B, C \in \left(0, \frac{\pi}{2}\right)$. Since $1+\tan ^{2} \theta=\sec ^{2} \theta$, the original inequality is equivalent to $\cos A \cos B \cos C(\cos A \sin B \sin C+\sin A \cos B \sin C+\sin A \sin B \cos C) \leqslant \frac{4}{9}$.

Because $\cos (A+B+C)=\cos A \cos B \cos C-(\cos A \sin B \sin C+$ $\sin A \cos B \sin C+\sin A \sin B \cos C)$, (1) can be transformed into
$$\cos A \cos B \cos C[\cos A \cos B \cos C-\cos (A+B+C)] \leqslant \frac{4}{9}$$

Let $\theta=\frac{A+B+C}{3}$, by the AM-GM inequality and Jensen's inequality ($y=\cos x$ is a concave function on $\left(0, \frac{\pi}{2}\right)$), we have
$$\cos A \cos B \cos C \leqslant\left(\frac{\cos A+\cos B+\cos C}{3}\right)^{3} \leqslant\left(\cos \frac{A+B+C}{3}\right)^{3}=\cos ^{3} \theta$$

To prove (2), it suffices to show
$$\cos ^{3} \theta\left(\cos ^{3} \theta-\cos 3 \theta\right) \leqslant \frac{4}{9}$$

Since $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$, (3) is equivalent to
$$\cos ^{4} \theta\left(1-\cos ^{2} \theta\right) \leqslant \frac{4}{27}$$

By the AM-GM inequality, we get
$$\begin{array}{l} \cos ^{4} \theta\left(1-\cos ^{2} \theta\right)=4 \cdot \frac{1}{2} \cos ^{2} \theta \cdot \frac{1}{2} \cos ^{2} \theta\left(1-\cos ^{2} \theta\right) \\ \leqslant 4 \cdot\left[\frac{\frac{1}{2} \cos ^{2} \theta+\frac{1}{2} \cos ^{2} \theta+\left(1-\cos ^{2} \theta\right)}{3}\right]^{3}=\frac{4}{27} \end{array}$$

Therefore, (3) holds. Hence, the original inequality is true.