Mathematics

The key is recognizing that $\cos A \cos B$ is bounded above by 1, and then finding the tightest constraint on the sum by treating $\cos C$ separately. Since all angles lie in $(-\frac{\pi}{2}, \frac{\pi}{2})$ where cosine is positive and decreasing, you can use the fact that $\cos A \cos B \le 1$ and $\cos C \le 1$—but that naive bound gives 2, which is too loose. The real insight is to use calculus or algebraic manipulation to show that the maximum of $\cos A \cos B + \cos C$ occurs at a specific configuration (like when $A = B = 0$ and $C = 0$), yielding exactly $\frac{3}{2}$.

The key is recognizing that $\cos A \cos B$ is bounded above by 1, and then finding the tightest constraint on the sum by treating $\cos C$ separately. Since all angles lie in $(-\frac{\pi}{2}, \frac{\pi}{2})$ where cosine is positive and decreasing, you can use the fact that $\cos A \cos B \le 1$ and $\cos C \le 1$—but that naive bound gives 2, which is too loose. The real insight is to use calculus or algebraic manipulation to show that the maximum of $\cos A \cos B + \cos C$ occurs at a specific configuration (like when $A = B = 0$ and $C = 0$), yielding exactly $\frac{3}{2}$.