# 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.

• If a triangle is not obtuse, then $a^2+b^2\le c^2$ for all permutations of $(a,b,c)$. Try using this fact coupled with the law of sines. Commented Jul 14, 2022 at 18:55
• @TheBestMagician that means $\sin^2C\ge\sin^2A+\sin^2B$? Commented Jul 14, 2022 at 19:05
• I made a typo, $a^2+b^2\ge c^2$ haha. But yeah applying law of sines you get $\sin A=\frac{a}{2R}$, etc. so $\sin^2 A+\sin^2 B\ge \sin^2 C$. Commented Jul 14, 2022 at 19:07
• @TheBestMagician oh yes. $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$ Since cosine is non-negative. So, $a^2+b^2\ge c^2$. Is there any other way to prove this fact? Commented Jul 14, 2022 at 19:10
• Honestly, I'm not sure either. Try using AM-GM. Commented Jul 14, 2022 at 19:21

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)!!

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

• Thank you ..... Commented Jul 15, 2022 at 2:59

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 $$\dfrac{12}{\pi}.$$

• Thanks. Any chance we could give a proof without partial derivatives? Because that's not in syllabus. Commented Jul 15, 2022 at 2:58