Estimate involving the exponential function 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!
 A: 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
A: 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


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....
A: 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 $0<\xi<x$.
A: 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.

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