Finding a bound for $\sum_{\text{cyc}}\frac{\sin B+\sin C}A$ if $\triangle ABC$ is not obtuse. The following question appeared in a JEE mock exam held two days ago.
Question:
$\triangle ABC$ is not obtuse then value of $\displaystyle\sum_{\text{cyc}}\frac{\sin B+\sin C}A$ must be greater than

*

*A) $\frac6\pi$

*B) $3$

*C) $\frac{12}\pi$

*D) $\frac1\pi$
My Attempt:
I first tried with sine rule $$\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}$$
But couldn't do anything with it.
Then I used $\sin B+\sin C=2\sin\left(\frac{B+C}2\right)\cos\left(\frac{B-C}2\right)=2\cos\frac A2\cos\left(\frac{B-C}2\right)$
But couldn't finish this approach either.
Then I tried using Jensen inequality but in vain.
Then I thought I would assume a function and find its minimum value. But couldn't decide what function to take.
 A: Another answer, elementary this time. Make everything have as common denominator $ABC$. Then the sum is
$BC\dfrac{\sin B+\sin C}{ABC}$+$AC\dfrac{\sin C+\sin A}{ABC}$+$AB\dfrac{\sin A+\sin B}{ABC}$ $(1)$
Now notice that the function $\,\,\dfrac{\sin x}{x}$ is strictly decreasing on $[0,\dfrac{\pi}{2}]$. (Elementary calculus). Therefore
$\dfrac{\sin A}{A}\geq\dfrac{2}{\pi}$. Likewise for $B,C$. Then $(1)$ gives that the sum is
$\geq$
$\dfrac{B}{A}$$\dfrac{2}{\pi}$+$\dfrac{C}{A}.\dfrac{2}{\pi}$+......$\geq\,6\dfrac{2}{\pi}$=$\dfrac{12}{\pi}$.
(Because $\dfrac{A}{B}+\dfrac{B}{A}\,\geq\,2$ for any positive $A,B.$)
This does NOT provide a minimum but it gives one of the answers (and therefore all the others are correct)!!
A: For an exam, I would try first a trivial case, $A=B=C=\frac\pi 3$. Then $\sin A=\frac{\sqrt 3}2$, so $$\sum_{cyc}\frac{\sin B+\sin C}{A}=3\frac{2\sin A}{A}=\frac{9\sqrt 3}\pi$$
The largest value from your choices is $\frac{12}{\pi}$, which is smaller than the value above. So the answer is C
A: A previous answer is correct, but is not a proof! There are many ways to get a minimum! The simplest is to set $A=x, B=y, \,\,C=\pi-x-y$ and try to minimize the function
$f(x,y)=\dfrac{siny+sin(\pi-x-y)}{x}+\dfrac{sin(\pi-x-y)+sinx}{y}+\dfrac{sinx+siny}{\pi-x-y}$.
Taking partial derivatives we get:
$\dfrac{\partial f}{\partial x}$= $-\dfrac{1}{x^{2}}(siny+sin(x+y))+\dfrac{1}{x}cos(x+y)+\dfrac{1}{y}(cos(x+y)+cosx)+\dfrac{cosx(\pi-x-y)+sinx+siny}{(\pi-x-y)^{2}}=0$
and likewise for $\,\,\dfrac{\partial f}{\partial y}$. Now it is time to use our intuition and see that $x=\dfrac{\pi}{3}$ and $y=\dfrac{\pi}{3}$ are solutions of the system and hence a stationary point.
Therefore the value of $f$ at this point is $\,\dfrac{9\sqrt{3}}{\pi}$
and to make sure it is not a maximum we use $A=\dfrac{\pi}{2}, B=\dfrac{\pi}{4}, C=\dfrac{\pi}{4}$ which gives $\dfrac{8+6\sqrt{2}}{\pi}$ a greater value than $\dfrac{9\sqrt{3}}{\pi}$. Therefore, the minimum value of the function is $\,\,\dfrac{9\sqrt{3}}{\pi}$ and hence all answers are correct. If we have to decide which value is closer to the minimum, then it is $\dfrac{12}{\pi}.$
