Question: **Context:** I've just got into reading papers but sometimes struggle a little bit with the "basics" i.e. the things that the authors take for granted.

**Problem:** I'd like to show that $$e^{-x} \leq 1-x+x^2 \textrm{ for all } x\geq 0. $$

**Ideas:** I tried using the taylor series of $e^{-x}$ which led me to $$e^{-x}=1-x+\frac {x^2} 2 +\sum\_{k=3}^\infty \frac{(-x)^k}{k!} $$ and now it remains to show that $$\sum\_{k=3}^\infty \frac{(-x)^k}{k!} \leq \frac {x^2}{2}.$$

This is where I'm stuck. Any help or hint is appreciated!

Answer: 
The statement is clearly true for $x=0$. If $x>0$, we have, by Taylor's theorem with Lagrange remainder,
$$
e^{ - x} = 1 - x + \frac{{x^2 }}{2}e^{ - \xi } < 1 - x + \frac{{x^2 }}{2}<1+x+x^2
$$
with a suitable $0<\xi<x$.
First do a drawing and you realize that

$$e^{-x}<1-x+x^2$$

$\forall x >0$

to prove it simply compare the derivatives of LHS vs RHS

---

[![enter image description here](https://i.stack.imgur.com/6nrDG.jpg)](https://i.stack.imgur.com/6nrDG.jpg)

If you compare the two derivatives you see that for all $x\_0 \in (0;0.5)$ the derivative of $e^{-x}$ is greater, in absolute value, w.r.t. the other one. when $x>0.5$ the parabola increases....