Geometric proof of $\sin x \geq x - x^3 /6 $?

You meant this holds for acute $x$. Take a radius-$\sqrt{2}$ centre-$O$ circle of radii $OA,\,OB$ with $\angle AOB=x$. We'll work in Cartesian coordinates $X,\,Y$; rotate the diagram so the line segment $AB$, which is of length $2\sqrt{2}\sin\frac{x}{2}$, has endpoints at $$X=\pm\sqrt{2}\sin\frac{x}{2},\,Y=0,$$and let $M$ be the midpoint of the arc $AB$. The circular segment has area $x-\sin x$, which we wish to prove $\le\frac{1}{6}x^3$. We can asymptotically approximate the arc as a parabola passing through the aforementioned endpoints and $$M=\left(0,\,\sqrt{2}\left(1-\cos\frac{x}{2}\right)\right)=\left(0,\,2\sqrt{2}\sin^2\frac{x}{4}\right).$$The parabola has equation $$Y=\frac{\sqrt{2}\sin^2\frac{x}{4}}{\sin^2\frac{x}{2}}\left(2\sin^2\frac{x}{2}-X^2\right),$$so the area under the parabola is $$\frac{2\sqrt{2}\sin^2\frac{x}{4}}{\sin^2\frac{x}{2}}\frac{4\sqrt{2}\sin^3\frac{x}{2}}{3}=\frac{16}{3}\sin^2\frac{x}{4}\sin\frac{x}{2}.$$(Admittedly I had to use $\int_0^a (a^2-X^2)dX=\frac{2a^3}{3}$ there, but geometrically this is equivalent to a pyramid's volume, so look up your favourite "classical" proof of that.) Asymptotically this is approximately, but less than, $$\frac{16x^3}{3\times 4^2\times 2}=\frac{x^3}{6}.$$

Here's a simple proof, not original by me, using calculus. It can easily be extended to show that the power series for sine and cosine are enveloping, that is successive terms bracket the function.

Start with this definition of since and cosine:

$\sin' = \cos $, $\cos' = -\sin $, $\sin(0) = 0$, $\cos(0) = 1$. These imply $\sin^2+\cos^2 = 1$.

For small $t$, $1 \ge \cos(t) \ge 0 $ so $\sin(x) =\int_0^x \cos(t)dt \le x $. Therefore $1-\cos(x) =\int_0^x \sin(t) dt \le \int_0^x t dt = \frac{t^2}{2} $ so $\cos(t) \ge 1-\frac{t^2}{2} $.

Therefore $\sin(x) =\int_0^x \cos(t)dt \ge\int_0^x (1-\frac{t^2}{2})dt =x-\frac{x^3}{6} $.

So we already have $x-\frac{x^3}{6} \le \sin(x) \le x $.