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: 
This sort of problems can be proved by construct a function as follows:
$$
g(x) = e^{-x} - 1 + x - x^2
$$
Just to prove $g(x) \le 0 $ for $x \ge 0$

you can discover that usually $g(0) = 0 , g'(0) = 0, g''(x) < 0,x \ge 0 $
by Monotonicity you can get what you want to prove
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 $\displaystyle\int\_a^x f(t)\,dt\ge0$.