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Prove that $$3x-x^3<\frac2{\sin2x},\forall x\in\left(0,\frac\pi2\right)$$

I have tried by proving that $$3x-x^3<\frac9{5\pi}x+\frac32<\frac2{\sin2x},\forall x\in\left(0,\frac\pi2\right)$$ with Jensen's inequality, but I hoped this problem would have a simpler solution.

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HINT:

$$\sin 2x\le1 \implies \frac{2}{\sin 2x}\ge 2$$

and

$$3x-x^3\le 2$$

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  • $\begingroup$ Thank you, I was foolish to see this. $\endgroup$
    – mja
    Sep 2 '15 at 16:16
  • $\begingroup$ You're welcome. My pleasure. And no you're not foolish. I have missed the tree in the forest many times. It is easy to do. You're fine! $\endgroup$
    – Mark Viola
    Sep 2 '15 at 16:17
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The range of the function $f(x)=3x-x^3$ for $x\in(0,\pi/2)$ is $f(x)\in(0,2)$.

For the function $g(x)=\dfrac2{\sin2x}$ for $x\in(0,\pi/2)$ the range is clearly $g(x)\in(2,\infty)$.

The inequality is quite clear now.

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  • $\begingroup$ Just like mine ;-)) $\endgroup$
    – Mark Viola
    Sep 2 '15 at 16:15
  • $\begingroup$ Yeah well the mobile site does not display answers in real-time. $\endgroup$
    – najayaz
    Sep 2 '15 at 16:16
  • $\begingroup$ No worry. I've posted answers like that many times. It's good to see alignment with our answers! $\endgroup$
    – Mark Viola
    Sep 2 '15 at 16:17
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Take $f(x)=(3x-x^3)\sin(2x)$ over $I=\left[0,\frac{\pi}{2}\right]$. It is a non-negative function since $\pi^2<12$. Since over the same interval we have $\sin(2x)\leq \frac{4}{\pi^2}(2x)(\pi-2x)$, it is enough to prove that:

$$ \forall x\in I,\qquad x^2(3-x^2)(\pi-2x)\leq \frac{\pi^2}{4}. \tag{1}$$ By differentiation we may locate the absolute maximum of $g(x)=x^2(3-x^2)(\pi-2x)$ over $I$ around $x=0.88552583$. In such a point $g(x)$ is about $2.38141<2.4674<\frac{\pi^2}{4}$, so $(1)$ holds.

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  • $\begingroup$ It's much simpler. $2\csc (2x)\ge 2$ and $3x-x^3\le 2$. $\endgroup$
    – Mark Viola
    Sep 2 '15 at 16:14
  • $\begingroup$ @Dr.MV: oh, nice. I thought it was possible to separate the two functions by putting a line in between (since they are a convex and a concave function), but I didn't try the most simple approach :D $\endgroup$ Sep 2 '15 at 16:34
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    $\begingroup$ I began thinking about ways to do this and then it just struck me. It's very easy to miss the tree from the forest so they say. $\endgroup$
    – Mark Viola
    Sep 2 '15 at 16:39
  • $\begingroup$ @Dr.MV "sometimes you can't see the wood for the trees" is the phrase that springs to mind $\endgroup$ Sep 2 '15 at 18:10
  • $\begingroup$ @DavidQuinn This is the saying that I was paraphrasing. $\endgroup$
    – Mark Viola
    Sep 2 '15 at 19:17

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