How do we show 1−cos⁡x≥x23 for |x|≤1? My first idea was to write1−cos⁡x=12|eix−1|2which is true...

For ${\displaystyle \left|x\right|\le 1}$ the Taylor series
${\displaystyle \cos \left(x\right)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots }$
is an alternating series with terms that decrease in absolute value. It follows that for these x
${\displaystyle \cos \left(x\right)\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}\le 1-\frac{x^2}{2!}+\frac{x^2}{4!}=1-\frac{11}{24}{x}^2}$
and therefore
${\displaystyle 1-\cos \left(x\right)\ge \frac{11}{24}{x}^2}$
That is an even better estimate since ${\displaystyle \frac{11}{24}>\frac{13}{}}$
The same approach can be used to show that ${\displaystyle 1-\cos \left(x\right)\ge \frac{x^2}{3}}$ holds on the larger interval [−2,2]:
${\displaystyle \cos \left(x\right)\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}\le 1-\frac{x^2}{2!}+\frac{4x^2}{4!}=1-\frac{1}{3}{x}^2}$
because the ${\displaystyle \frac{x^{2n}}{\left(2n\right)!}}$ terms decrease in absolute value for ${\displaystyle n\ge 1}$

Since sinx is concave for ${\displaystyle x\in [0,\frac{\pi }{2}]}$, if ${\displaystyle x\in [0,1]}$ then ${\displaystyle \sin x\le x\sin 1\Rightarrow 1-\cos x\ge x^2\frac{12}{\sin 1}}$. The ${\displaystyle x^2}$ coefficient approximates 0.42. The generalization ${\displaystyle x\in [-1,1]}$ follows from ${\displaystyle 1-\cos x,c{x}^2}$ being even.

$f(x)=\frac{1-\cos x}{x^2}$ is an even function, so we only look at $x\in [0,1]$. It's easy to prove f(x) is continuous and differentiable at x=0, and
$f\prime (x)=\frac{x\sin x+2\cos x-2}{x^3}$
Now
$x\sin x+2\cos x-2\le 2\tan \frac{x}{2}\cdot \sin x-4{\sin }^2\frac{x}{2}$
$=2\tan \frac{x}{2}\cdot 2\sin \frac{x}{2}\cos \frac{x}{2}-4{\sin }^2\frac{x}{2}=0,\mathrm{\forall }x\in [0,1]$
Hence $f(x)\ge f(1)=1-\cos 1\approx .46>\frac{1}{3},1-\cos x\ge (1-\cos 1)x^2>\frac{1}{3}{x}^2$