Better proof of a numerical inequality of $e^x$ The inequality is

$$ e^z \leq 1+z+\frac{z^2/2}{1-|z|/3} \text{ for } |z|<3$$

I proved it by splitting it into 3 cases: $-3<z<0$, $z=0$ and $0<z<3$.
For $z=0$, both sides equals.
The other 2 cases are done with calculus. Define $f(x)=e^x-1-x-\frac{x^2/2}{1-|x|/3}$ and then replace $|x|$ by $x$ or $-x$ accordingly. Then just check the derivatives.
But in my opinion, it is sort of brute force, so I am wondering if there is faster(smarter) way to show it.

Note that, if $|z|<3$,\begin{align}e^z-1-z&=\frac{z^2}2+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots\\&=\frac{z^2}2\left(1+\frac z3+\frac{z^2}{3\times4}+\frac{z^3}{3\times4\times5}+\cdots\right)\\&\leqslant\frac{z^2}2\left(1+\frac{|z|}3+\frac{|z|^2}{3\times4}+\frac{|z|^3}{3\times4\times5}+\cdots\right)\\&\leqslant\frac{z^2}2\left(1+\frac{|z|}3+\frac{|z|^2}{3^2}+\frac{|z|^3}{3^3}+\cdots\right)\\&=\frac{z^2}2\cdot\frac1{1-|z|/3}.\end{align}
