How do I prove this trigonometric inequality?

Question

If $A,B,C \in (0,\frac{\pi}{2})$. Then prove that $$\frac{\sin(A+B+C)}{\sin(A)+\sin(B)+\sin(C)} < 1$$

Answer 1

Expand using the sum formula for sine: \begin{align} \sin(A+B+C)&=\sin(A+B)\cos(C)+\sin(C)\cos(A+B) \\ &=\sin(A)\cos(B)\cos(C)+\sin(B)\cos(A)\cos(C)+\sin(C)\cos(A+B) \end{align} Now since $A,B,C\in\left(0,\frac{\pi}{2}\right)$, we know $\max\{\cos(A),\cos(B),\cos(C),\cos(A+B)\}<1$, and therefore $$\sin(A)\cos(B)\cos(C)+\sin(B)\cos(A)\cos(C)+\sin(C)\cos(A+B)<\sin(A)+\sin(B)+\sin(C).$$ Combining what we have so far, we see that $\sin(A+B+C)<\sin(A)+\sin(B)+\sin(C)$, and since the left hand side cannot be zero, we can divide by it to obtain the result.

Answer 2

If $A+B+C\geq\pi$ our inequality is obviously true.

Let $A+B+C<\pi$ and $A\geq B\geq C$.

Since $(A+B+C,0,0)\succ(A,B,C)$ and $\sin$ is a concave function on $[0,\pi]$,

by Karamata we obtain: $$\sin(A+B+C)+\sin0+\sin0\leq\sin{A}+\sin{B}+\sin{C}$$ and we are done!