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

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    How about using Jensen's Inequality? – 2017-02-07
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    @ShreyAryan Is $\sin$ convex on $(0,\frac{\pi}{2})$? – 2017-02-07
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    No it isin't...But we can reverse the direction of the inequality. – 2017-02-07

2 Answers 2

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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!

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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.

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    But product of cosBcosC can exceed value 1, can't it? – 2017-02-07
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    @TaylorTed No. Their sum can, but not their product. – 2017-02-07
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    @TaylorTed No problem. – 2017-02-07