Geometric proof of $\sin x \geq x - x^3 /6 $? We know (from Taylor expansion for example) that if $x \geq 0$, then $\sin x \geq x - \frac{x^3}{6}$.
In Prove that: $\sin(x) \cos(x) \geq x-x^3$ a geometric proof of the inequality $\sin x \geq x - \frac{x^3}{4}$ is given. Is there any geometric proof of the first one (which is slightly stronger ?

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, $$\frac{16x^3}{3\times 4^2\times 2}=\frac{x^3}{6}.$$
