Problem 1282

$$\begin{array}{l} \sin A+\sin B+\sin C-\cos A-\cos B-\cos C \\ =(\sin A+\sin B)-(\cos A+\cos B)+(\sin C-\cos C) \\ =2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}-2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}+2 \sin \frac{C}{2} \sin \frac{C}{2}-2 \cos ^{2} \frac{C}{2}+1 \\ =1+2 \cos \frac{A-B}{2}\left(\sin \frac{A+B}{2}-\cos \frac{A+B}{2}\right)+2 \cos \frac{C}{2}\left(\sin \frac{C}{2}-\cos \frac{C}{2}\right) \\ =1+2 \cos \frac{A-B}{2}\left(\cos \frac{C}{2}-\sin \frac{C}{2}\right)+2 \cos \frac{C}{2}\left(\sin \frac{C}{2}-\cos \frac{C}{2}\right) \\ =1+2\left(\sin \frac{C}{2}-\cos \frac{C}{2}\right)\left(\cos \frac{C}{2}-\cos \frac{A-B}{2}\right) \\ =1+2 \sqrt{2} \sin \left(\frac{C}{2}-\frac{\pi}{4}\right) \cdot(-2) \sin \frac{A+C-B}{4} \sin \frac{B+C-A}{4} \\ =1+4 \sqrt{2} \sin \left(\frac{\pi}{4}-\frac{C}{2}\right) \sin \frac{A+C-B}{2} \sin \frac{B+C-A}{2}, \end{array}$$

Since $A, B, C \in \left(0, \frac{\pi}{2}\right)$, we have $0 < \sin \left(\frac{\pi}{4} - \frac{C}{2}\right) < 1$, $\sin \frac{A+C-B}{4} > 0$, and $\sin \frac{B+C-A}{4} > 0$,
thus $\sin A + \sin B + \sin C - \cos A - \cos B - \cos C > 1 > 0$.
From the proof process, we obtain a new problem:
In an acute $\triangle ABC$, prove that $\sin A + \sin B + \sin C > 1 + \cos A + \cos B + \cos C$.
This is a stronger statement than the original problem.
Furthermore, since $\cos A + \cos B + \cos C \in \left(1, \frac{3}{2}\right]$, we can derive:
In an acute $\triangle ABC$, prove that $\sin A + \sin B + \sin C > 2$.
This looks simpler and when $A \rightarrow \frac{\pi}{2}$, $B \rightarrow \frac{\pi}{2}$, $\sin A + \sin B + \sin C \rightarrow 2$, meaning 2 is the best constant.
How to prove it? We can use a construction method or other approaches.
Method 1: Utilize the function $y = \frac{\sin x}{x}$ which is monotonically decreasing in $\left(0, \frac{\pi}{2}\right)$, see Example 4 in 81.1.
Method 2 (Adjustment Method): Let $f(A, B, C) = \sin A + \sin B + \sin C$, we will prove:
$f(A, B, C) > 2$.
Consider the function $f(A, B, C)$ and its partial derivatives. We have:
$$\frac{\partial f}{\partial A} = \cos A, \quad \frac{\partial f}{\partial B} = \cos B, \quad \frac{\partial f}{\partial C} = \cos C.$$
Since $\cos A, \cos B, \cos C > 0$ in an acute triangle, $f(A, B, C)$ is an increasing function.
We also have:
$$\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2 \cos A \cos B \cos C > 2.$$

A new question arises: In an acute $\triangle ABC$, which is larger, $\sin^2 A + \sin^2 B + \sin^2 C$ or $1 + \cos A + \cos B + \cos C$?
When $(A, B, C) \rightarrow \left(\frac{\pi}{2}, \frac{\pi}{2}, 0\right)$, $\sin^2 A + \sin^2 B + \sin^2 C \rightarrow 2$, $1 + \cos A + \cos B + \cos C \rightarrow 2$,
When $(A, B, C) \rightarrow \left(\frac{\pi}{3}, \frac{\pi}{3}, \frac{\pi}{3}\right)$, $\sin^2 A + \sin^2 B + \sin^2 C = \frac{9}{4}$, $1 + \cos A + \cos B + \cos C = \frac{5}{2}$.
We conjecture that in an acute $\triangle ABC$, $\sin^2 A + \sin^2 B + \sin^2 C > 1 + \cos A + \cos B + \cos C$.
This is equivalent to:
$$\sin^2 A + \sin^2 B + \sin^2 C > 1 + \cos A + \cos B + \cos C$$
$$\sin^2 A + \sin^2 B + \sin^2 C > 2$$
$$2 + 2 \cos A \cos B \cos C > 2$$
$$2(\cos A + \cos B + \cos C) + \cos 2A + \cos 2B + \cos 2C > 1$$
$$(\cos A + \cos B + \cos C - 1) + (\cos A + \cos B + \cos C + \cos 2A + \cos 2B + \cos 2C) > 0$$
Since $\cos A + \frac{1}{2}(\cos 2B + \cos 2C) = \cos A + \cos (B+C) \cos (B-C) = \cos A [1 - \cos (B-C)] > 0$,
Similarly, $\cos B + \frac{1}{2}(\cos 2C + \cos 2A) > 0$, $\cos C + \frac{1}{2}(\cos 2A + \cos 2B) > 0$.
Thus, $\cos A + \cos B + \cos C + \cos 2A + \cos 2B + \cos 2C > 0$,
and $\cos A + \cos B + \cos C > 1$.
Therefore, $\sin^2 A + \sin^2 B + \sin^2 C > 1 + \cos A + \cos B + \cos C > 2$.

Note 1: In this example, we should not neglect or underestimate some seemingly clumsy methods, as they are often the closest to the essence, embodying the idea that "great clumsiness is clever, great foolishness is wise."
Note 2: For the inequality $\sin A + \sin B + \sin C > 2$, we provide another geometric approach.
Let the sides of $\triangle ABC$ be $a, b, c$, and the circumradius be $R$. The inequality to be proven can be transformed into $a + b + c > 4R$.
Construct $\triangle ABC$ and its circumcircle. Assume $a = BC$ is the largest side, and translate $\triangle ABC$ and its circumcircle to $\triangle A_1B_1C_1$ and its circumcircle, such that $B_1$ coincides with $C$,
and $C_1$ lies on the extension of $BC$, as shown in Figure 1-17. Let the other intersection of the two circles be $D$,
then $\angle BDC = \angle C_1DC$, and $BC = CC_1$, so $DB = DC_1$, $DC \perp BC_1$,
$DB$ and $DC_1$ are diameters of the two circles. Since $\triangle ABC$ is an acute triangle, $D$ lies inside the quadrilateral $AB C_1 A_1$,
thus $AB + AA_1 + A_1C_1 > DB + DC_1$, i.e., $a + b + c > 4R$.
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