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Thread: Trigonometric inequality in a triangle?

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

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

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

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

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

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

    *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 $\displaystyle \approx 1.94$ It could be possible to approximate $\displaystyle A_{max}=f(B)\pm\Delta$ If $\displaystyle \Delta$ 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 $\displaystyle \sin A\sin 2A+\sin(A+B)\sin(2A+2B)=0$ will be a stationary point in the original formula. Substitute A=M-R and B=2R and the equation becomes somewhat manageable: $\displaystyle \sin (M-R)\sin (2M-2R)+\sin(M+R)\sin(2M+2R)=0$

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

    So the stationary points of $\displaystyle f(A,B)=\sin^3A+\sin^3B+\sin^3(A+B)$ are precisely $\displaystyle (A,B)=(M+R,2R)$ for $\displaystyle R=\sin^{-1}(\sin M\sqrt{4\sin^2M-1})$ . 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 $\displaystyle (0,\frac\pi2]^2$, 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, $\displaystyle (M+R,2R)=(2R,M+R)$ ~ $\displaystyle M+R=2R$ ~ $\displaystyle M=R=\sin^{-1}(\sin M\sqrt{4\sin^2M-1})$. Solving, $\displaystyle M=\frac\pi4$, so $\displaystyle (A,B)=M=(\frac\pi2,\frac\pi2)$.

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

    QED
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  4. #4
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    We are stupid

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