Prove that: $\sin(x) \cos(x) \geq x-x^3$

A Geometric Proof

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

Consider the wedge of the unit circle below:

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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.

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.