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Prove that the following inequality holds for $x\ge0$ :

$\sin(x) \cos(x) \geq x-x^3$

This is an inequality often met during my high school classes I also used for this problem yesterday. I'm interested in a non-calculus proof if this is possible.

Proof involving calculus:

Let's consider

$f(x) = \sin(x) \cos(x)-x+x^3$ then $f'(x) = 3 x^2-2\sin^2(x)\tag1$ $x\ge \sin(x)\tag2$ From $(1)$ and $(2)$ we immediately notice that $f'(x)\ge0$ and taking into account that $f(0)=0$ we may conclude that the inequality holds. Thanks.

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    Maybe it helps that from \sin x and the fact that for 0 we have \sin x>0 we get \sin^2 x (for 0, we can just square the inequality, and beyond we just notice that \sin^2 x\le 1 < x). Therefore we get \cos^2 x = 1-\sin^2 x > 1-x^2 and thus x\,\cos^2 x > x-x^3. However I have no idea how to prove without calculus that x\,\cos x < \sin x for x<\pi/2 (you can see that by plotting both functions).2012-08-23

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I can't seem to come up with a non-calculus method of solving, but taking @dato's trig trick, you can multiply both sides by $2$ yielding:

$2\sin x\cos x\geq 2(x-x^3)$

$\sin2x\geq 2x-2x^3$

And using a Taylor expansion:

$2x-\frac{8x^3}{3!}+\mathcal{O}(x^5)\geq 2x-2x^3$ $\frac{12x^3}{6}-\frac{8x^3}{6}+\mathcal{O}(x^5) = \frac{4x^3}{6}+\mathcal{O}(x^5)\geq 0$ ... when $x\geq 0$.

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    Once again, and despite the upvotes, one should mention that limited expansions CANNOT yield such inequalities, for any fixed $x\gt0$.2012-08-23
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A Geometric Proof

The posed inequality is equivalent to $\sin(x)\ge x-x^3/4.$

Consider the wedge of the unit circle below:

$\hspace{32mm}$enter image description here

The area of the whole wedge (red and green regions) is $\frac12x$, and the area of the green triangle is $\frac12\sin(x)$. Thus, we get that $\sin(x)\le x$. Furthermore, the area of the red region is $\frac12(x-\sin(x))$.

Noting that the red region is contained in the rectangle with base $2\sin(x/2)$ and height $1-\cos(x/2)$, we get that $ \begin{align} \tfrac12(x-\sin(x)) &\le2\sin(x/2)(1-\cos(x/2))\\ &=4\sin(x/2)\sin^2(x/4)\\ &\le x^3/8 \end{align} $ which yields $ x-x^3/4\le\sin(x) $ as desired.

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    @ robjohn: Intaglio seems nice, but I don't have Mac. I'll try Mathematica. Thanks for information!2012-09-08
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This is a completely revamped proof. We only use the two inequalities $ x \geq \sin(x) \text{ and } \tan(x) \geq x $ which have elementary proofs here.

Case 1: $ 0 \leq x < \pi/2$

We know that $x \geq \sin(x)$. This gives us $x^2 \geq \sin^2(x)$, for with $x \leq \pi$ both terms are positive and simply squaring is justified, and with $x \geq \pi$, $x^2 \geq \pi^2 \geq 1 \geq \sin^2(x)$.

We can do some manipulations on this inequality to find an inequality involving cos: $ x^2 \geq \sin^2(x)\\ x^2 \geq 1 - \cos^2(x)\\ \cos^2(x) \geq 1- x^2 $ We now have $ \sin(x) \cos(x) = \frac{\sin(x)}{\cos(x)} \cos^2(x) = \tan(x) \cos^2(x) \geq x(1- x^2) = x-x^3 $ as required.

Case 2: $ \pi/2 \leq x$

The previous argument only works for small $x$, where tan is well behaved. For $ \pi/2 \leq x$ we see that $x-x^3 \leq \pi/2 - (\pi/2)^2 \leq -1$ (as the function $g(x) = x-x^3$ is decreasing for $x \geq 1$), and as $-1 \leq \sin(x) \cos(x)$ (as both sin and cos are large than -1), the result follows.

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    Nice answer (+1)! Both of these inequalities, $\sin(x)\le x\le\tan(x)$, are shown in [this geometric proof](http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1/75151#75151), too.2012-08-23
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As others have already noticed, multiplying both sides by $2$ makes it sufficient to prove $\sin 2x \ge 2x - 2x^3$ or $\sin x \ge x - \frac{x^3}{4}$.

One method of proving this is the following:

It is well known that $\sin x \le x$, therefore

$\int_0^x (t - \sin t) \, dt \ge 0,$

giving us that $\cos x \ge 1 - \frac{x^2}{2}$.

Applying the same thing again, we have

$\int_0^x \cos t \, dt \ge \int_0^x \left(1 - \frac{t^2}{2}\right) \, dt$

giving us that $\sin x \ge x - \frac{x^3}{6}$, so in fact we've proven a stronger inequality!

Note: It is easy to see that this method yields easily that the Taylor's series stopped at positive terms (negative terms) overestimates (respectively, underestimates) $\sin x, \cos x$.

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    Well, a positive function having a positive area below it seemed much more intuitive than anything else I could think of..2012-08-23
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Multiply both sides by $2$ and then by drawing curves of the two functions, namely $\sin 2x$ and $x^3-x$, find the region where the curve of $\sin x$ is above the other in the graph.

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    @MichaelHardy: Well, seems I forgot a dollar sign and it was a bit messed up.2012-08-23