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Prove that for any triangle $ABC$, have $\sin{A}+\sin{B}+\sin{C} \leq \dfrac{3\sqrt{3}}{2}$.

I know that we have to use Jensen's inequality here, but I am not sure how to apply it. We know that $A+B+C = 180^{\circ}$, so seeing that sine is concave on $[0,\pi]$, we can let $f(x) = \sin{x}$.

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    $\begingroup$ $\frac{\sin A+\sin B+\sin C}{3}\le \sin\left(\frac{A+B+C}{3}\right)=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$. $\endgroup$
    – user236182
    Jan 26, 2016 at 22:28
  • $\begingroup$ What did you try? $\endgroup$
    – sinbadh
    Jan 26, 2016 at 22:28

1 Answer 1

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For a concave function, such as sine on $[0,\pi]$, we have $$\sin\Bigl(\frac{A+B+C}3\Bigr)=\sin\frac\pi3\le\frac13(\sin A+\sin B+\sin C),$$ whence the required inequality.

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