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Math Help - Inequalities(3):

  1. #1
    Member great_math's Avatar
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    Inequalities(3):

    For any acute angle \Delta ABC , prove that

    \frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}\le\frac{9\sqrt3}{2\pi}
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Using Jensen's inequality for the function \frac{sin x}{x},

    \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{sin \frac{A+B+C}{3}}{\frac{A+B+C}{3}}

    \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{sin \frac{\pi}{3}}{\frac{\pi}{3}}

    \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{\frac{\sqrt{3}}{2}}{\frac{\pi}{3}}

    \frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C} \leq 3 \frac{\frac{\sqrt{3}}{2}}{\frac{\pi}{3}}

    \frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C} \leq \frac{9\sqrt3}{2\pi}
    Last edited by alexmahone; June 19th 2009 at 05:48 AM.
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  3. #3
    Super Member malaygoel's Avatar
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    Please check the question....I think it is incorrect.

    .........my mistake ::::Confused A and a

    How to delete the post???
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  4. #4
    Senior Member pankaj's Avatar
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    Let f(x)=\frac{\sin x}{x}

    f'(x)=\frac{x\cos x-\sin x}{x^2}=\frac{\cos x(x-\tan x)}{x^2}<0,since x<\tan x for 0<x<\frac{\pi}{2}

    f''(x)=\frac{-x^2\sin x-2x\cos x+2\sin x}{x^3}

    Let g(x)=-x^2\sin x-2x\cos x+2\sin x

    g'(x)=-x^2\cos x<0 for 0<x<\frac{\pi}{2}

    For x>0,we have g(x)<g(0) and thus g(x)<0

    Therefore, f''(x)<0 for all x\in \left(0,\frac{\pi}{2}\right).

    Thus the function y=f(x) is concave downwards for all x\in \left(0,\frac{\pi}{2}\right).

    \therefore f\left(\frac{A+B+C}{3}\right)\geq \frac{f(A)+f(B)+f(C)}{3}

    \frac{\sin\left(\frac{A+B+C}{3}\right)}{\frac{A+B+  C}{3}}\geq \frac{\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}}{3}

     <br />
\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}\leq \frac{9\sqrt{3}}{2\pi}<br />
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  5. #5
    Senior Member pankaj's Avatar
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    Quote Originally Posted by alexmahone View Post
    Using Jensen's inequality for the function \frac{sin x}{x},

    \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{sin \frac{A+B+C}{3}}{\frac{A+B+C}{3}}
    \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{sin \frac{\pi}{3}}{\frac{\pi}{3}}
    \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{\frac{\sqrt{3}}{2}}{\frac{\pi}{3}}
    You first of all need to prove that f''(x)<0, \forall x\in\left(0,\frac{\pi}{2}\right) to use the Jensen's inequality
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by pankaj View Post
    You first of all need to prove that f''(x)<0, \forall x\in\left(0,\frac{\pi}{2}\right) to use the Jensen's inequality
    I know that, lol.
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  7. #7
    Super Member malaygoel's Avatar
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    Quote Originally Posted by pankaj View Post

    Thus the function y=f(x) is concave downwards for all x\in \left(0,\frac{\pi}{2}\right).

    \therefore f\left(\frac{A+B+C}{3}\right)\geq \frac{f(A)+f(B)+f(C)}{3}

    How inequality follows from the condition?
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  8. #8
    Senior Member pankaj's Avatar
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    Quote Originally Posted by malaygoel View Post
    How inequality follows from the condition?
    Consider the three points on the curve (A,f(A)) (B,f(B)) and (C,f(C)). When joined they form a triangle which obviously lies below the curve y=f(x).Now centroid of this triangle is inside the triangle and thus centroid is also below the curve.

    Co-ordinates of centroid=  \left(\frac{A+B+C}{3},\frac{f(A)+f(B)+f(C)}{3}\rig  ht).

    The corresponding point on the curve with the same x-coordinate will be \left(\frac{A+B+C}{3},f(\frac{A+B+C}{3})\right)

    Since this point is above the centroid .Therefore, f(\frac{A+B+C}{3})>\frac{f(A)+f(B)+f(C)}{3}.

    Hey,you can try plotting y=\ln x and draw a triangle and check for yourself
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  9. #9
    Super Member malaygoel's Avatar
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    Quote Originally Posted by great_math View Post
    For any acute angle \Delta ABC , prove that

    \frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}\le\frac{9\sqrt3}{2\pi}
    Does the information that triangle is acute is of any help here? The proofs seem to be valid for obtuse triangles as well.
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