Prove that: $\sin(x) \cos(x) \geq x-x^3$ 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.
 A: A Geometric Proof
The posed inequality is equivalent to $\sin(x)\ge x-x^3/4.$
Consider the wedge of the unit circle below:
$\hspace{32mm}$
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.
A: 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.
A: 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) $\sin x, \cos x$.
A: 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.
A: 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 $x\geq 0$.
