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in $\Delta ABC$, if $A,B,C\in (0,\pi/2]$,show that $$\sin{A}+\sin{B}+\sin{C}>2$$

This problem have many nice methods? Thank you

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It seems that it follows from Jordan's inequality. $$\frac{2}{\pi}(A + B +C) = 2 \le \sin(A) + \sin (B) + \sin (C)$$

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If $A,B$ and $C$ are angles of a triangle, then $\cos A$,$\cos B$ and $\cos C$ are roots of following polynomial: $$ 4R^2x^3-4R(R+r)x^2+(p^2+r^2-4R^2)x+(2R+r)^2-p^2=0 $$ Then we get: $$ \cos A \cos B \cos C=\frac{p^2-(2R+r)^2}{4R^2} $$

If all angles of the triangle $ABC$ is less than $\frac{\pi}{2}$, then all $\cos$ functions are positive and we have: $$ p\geq 2R+r \implies a+b+c\geq 4R+2r $$ Then using $a=2R\sin A$,$b=2R\sin B$ and $c=2R\sin C$, we get: $$ \sin A+\sin B+\sin C\geq 2+\frac{r}{R}\geq 2 $$

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$\sin$ is a concave function on $\left(0,\frac{\pi}{2}\right]$ and $\left(\frac{\pi}{2},\frac{\pi}{2},0\right)\succ(\alpha,\beta,\gamma)$.

Thus, by Karamata $$\sin\alpha+\sin\beta+\sin\beta>\sin\frac{\pi}{2}+\sin\frac{\pi}{2}+\sin0=2.$$ Done!

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