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Math Help - Trigonometric inequality in a triangle?

  1. #1
    Super Member fardeen_gen's Avatar
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    Trigonometric inequality in a triangle?

    In a \Delta ABC, prove that:
    \sin^{3}A+\sin^{3}B+\sin^{3}C < 2
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  2. #2
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    Getting started...

    Well, to start, \sin^{3}A+\sin^{3}B+\sin^{3}C=\sin^{3}A+\sin^{3}B+  \sin^{3}(\pi-A-B)=\sin^{3}A+\sin^{3}B+\sin^{3}(A+B)

    Now find the maximum of f(A,B) within the square (0,\frac\pi2)^2

    WLOG, \frac{\partial f}{\partial A}=3\sin^2A\cos A+3\sin^2(A+B)\cos(A+B)=0

    Or \sin A\sin 2A+\sin(A+B)\sin(2A+2B)=0

    *Separating A and B is eluding me, and it might not even be possible. The maxima of A with respect to B seem to always be \approx 1.94 It could be possible to approximate A_{max}=f(B)\pm\Delta If \Delta is sufficiently small, this will prove the hypothesis, but it's the ugly way to go about it.
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  3. #3
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    Picking up...

    Picking up where I left off, WLOG, the choice of A,B satisfying \sin A\sin 2A+\sin(A+B)\sin(2A+2B)=0 will be a stationary point in the original formula. Substitute A=M-R and B=2R and the equation becomes somewhat manageable: \sin (M-R)\sin (2M-2R)+\sin(M+R)\sin(2M+2R)=0

    Solving, R=\sin^{-1}(\sin M\sqrt{4\sin^2M-1}) for |\sin M|\geq\frac{\sqrt{3}}3

    So the stationary points of f(A,B)=\sin^3A+\sin^3B+\sin^3(A+B) are precisely (A,B)=(M+R,2R) for R=\sin^{-1}(\sin M\sqrt{4\sin^2M-1}) . This parametric equation over M traces a line in the AB plane directly under the peaks (or valleys) of the 3-dimensional f above.

    Now remember, we found the partial with respect to A, but by symmetry, the partial w.r.t. B is the exact same, simply switching A and B. Thus, We have two parametric lines running across the AB plane in the square (0,\frac\pi2]^2, one exactly under the North-South ridges/valleys, and one tracing the East-West ridges/valleys. Their intersection point will be a point at which the tangent plane is perfectly horizontal. So, (M+R,2R)=(2R,M+R) ~ M+R=2R ~ M=R=\sin^{-1}(\sin M\sqrt{4\sin^2M-1}). Solving, M=\frac\pi4, so (A,B)=M=(\frac\pi2,\frac\pi2).

    Now, \frac{\partial^2f}{\partial A^2}(\frac\pi2,\frac\pi2)=-3 (maximum), and \frac{\partial^2f}{\partial B\partial A}(\frac\pi2,\frac\pi2)=0 (no info? another way to determine?). Therefore, f(\frac\pi2,\frac\pi2)=2 is the highest point on the function in (0,\frac\pi2]^2. Since no legal triangle can have two right angles, WLOG, for all 0<A\leq B\leq \frac\pi2, \sin^3A+\sin^3B+\sin^3(A+B)<2

    QED
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  4. #4
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    We are stupid

    Consider this: \lim_{\epsilon\rightarrow0} ~\sin^n(90^\circ-\epsilon)+\sin^n(90^\circ)+\sin^n(\epsilon) = 2 Looks like fardeen_gen gets the last laugh... again.
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