Sines

Prove that in any acute angled triangle ABCABC , we have sin⁡AA+sin⁡BB+sin⁡CC>6π\large \frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} > \frac{6}{\pi}

#Geometry

Note by Vilakshan Gupta
3 years, 6 months ago

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Comments

sin⁡Aa=sin⁡Bb=sin⁡Cc=12R(Law of sines)R is the radius of the circumcircle of the trianglesin⁡Aa+sin⁡Bb+sin⁡Cc=32RBy scaling the same triangle we can obtain R as large or small as needed.Thus there is no clear maximum or minimum for the given equation.\begin{aligned} \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} &=\frac{1}{2R}\hspace{5mm}\color{#3D99F6} \text{(Law of sines)}\\ \text{R is the radius of the circumcircle} &\text{ of the triangle} \\ \frac{\sin A}{a} + \frac{\sin B}{b} + \frac{\sin C}{c}&=\frac{3}{2R}\\\\ \text{By scaling the same triangle we can} &\text{ obtain R as large or small as needed.}\\ \text{Thus there is no clear maximum or} &\text{ minimum for the given equation.} \end{aligned}

are you sure the intended equation isn't

sin⁡AA+sin⁡BB+sin⁡CC>6π\large \frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C}>\frac{6}{\pi} ?

Anirudh Sreekumar - 3 years, 6 months ago

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My Bad. Actually in denominator, there should be angles and not sides. Yes you are saying correct. Please help me with this,

Vilakshan Gupta - 3 years, 6 months ago

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Let,f(x)=sin⁡xxx∈(0,π2]f′(x)=xcos⁡x−sin⁡xx2(1)Let,g(x)=xcos⁡x−sin⁡xg(0)=0(2)g′(x)=−xsin⁡x  ⟹  g(x) is decreasing for x∈(0,π2](3)From (2) and (3) we have,g(x)<0 for x∈(0,π2](4)f′(x)=g(x)x2  ⟹  f′(x)<0 for x∈(0,π2] from (4)  ⟹  f(x)=sin⁡xx is decreasing for x∈(0,π2](5)Since ΔABC is acute 0<∠A,∠B,∠C<π2From (5) we have,sin⁡AA>sin⁡(π2)π2  ⟹  sin⁡AA>2π(6)Similarly,sin⁡BB>2π(7)sin⁡CC>2π(8)(6)+(7)+(8) givessin⁡AA+sin⁡BB+sin⁡CC>6π\begin{aligned}\text{Let,} f(x)&=\dfrac{\sin x}{x}\color{#3D99F6}\hspace{5mm} x\in\left({0,\dfrac{\pi}{2}}\right]\\ f'(x)&=\dfrac{x\cos x-\sin x}{x^2}\hspace{4mm}\color{#3D99F6}(1)\\\\ \text{Let,} g(x)&=x \cos x-\sin x\\ g(0)&=0\hspace{4mm}\color{#3D99F6}(2)\\ g'(x)&=-x\sin x\\ \implies g(x) &\text{ is decreasing for }x\in\left({0,\dfrac{\pi}{2}}\right] \hspace{4mm}\color{#3D99F6}(3)\\ \text{From } &\color{#3D99F6}(2)\color{#333333} \text{ and } \color{#3D99F6}(3)\color{#333333} \text{ we have,}\\ g(x)&<0 \text{ for }x\in\left({0,\dfrac{\pi}{2}}\right] \hspace{4mm}\color{#3D99F6}(4) \\ f'(x)&=\dfrac{g(x)}{x^2}\\ \implies f'(x)&<0 \text{ for }x\in\left({0,\dfrac{\pi}{2}}\right] \hspace{4mm}\color{#3D99F6}\text{ from }(4) \\ \implies f(x)&=\dfrac{\sin x}{x}\text{ is decreasing for }x\in\left({0,\dfrac{\pi}{2}}\right]\hspace{4mm}\color{#3D99F6}(5)\\\\ \text{Since }\Delta ABC \text{ is acute }\\ 0< \angle A,\angle B,\angle C<\dfrac{\pi}{2} \\ \text{From }\color{#3D99F6}(5) \color{#333333} \text{ we have,}\\ \dfrac{\sin A}{A}&>\dfrac{\sin (\dfrac{\pi}{2})}{\dfrac{\pi}{2}}\\ \implies\dfrac{\sin A}{A}&>\dfrac{2}{\pi}\hspace{4mm}\color{#3D99F6}(6)\\ \text{Similarly,}\\ \dfrac{\sin B}{B}&>\dfrac{2}{\pi}\hspace{4mm}\color{#3D99F6}(7)\\ \dfrac{\sin C}{C}&>\dfrac{2}{\pi}\hspace{4mm}\color{#3D99F6}(8)\\ \color{#3D99F6} (6)+(7)+(8) \text{ gives}\\ \dfrac{\sin A}{A}+\dfrac{\sin B}{B}+\dfrac{\sin C}{C}&>\dfrac{6}{\pi}\end{aligned}

Anirudh Sreekumar - 3 years, 6 months ago

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@Anirudh Sreekumar – Thanks.

Vilakshan Gupta - 3 years, 6 months ago

Please help me anyone. @Chew-Seong Cheong , @Brian Charlesworth , @Jon Haussmann , @Anirudh Sreekumar , @Hosam Hajjir , @Pi Han Goh

Vilakshan Gupta - 3 years, 6 months ago
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