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: 
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

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[![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....
For $x\ge0$, $$1-e^{-x}\ge0$$

Integrate twice from $0$ to $x$:

$$x+e^{-x}-1\ge0$$

$$\frac{x^2}2-e^{-x}+1-x\ge0.$$

You can continue forever.

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**Justification:**

If $f(x)\ge0$ from $x=a$, $\displaystyle\int f(x)\,dx$ is a growing function from $x=a$ and $\displaystyle\int\_a^x f(t)\,dt\ge0$.