<s>[INST] Proof that $\exp(x)-1 > x^2 ,\forall x>0$ I'm trying to prove that
$\exp(x)-1 > x^2 ,\forall x>0$
What I've tried so far:
for $0<x<1$ I know that the equation is valid because:
$\exp(x)-1= x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$
so considering only the first term ($x$) I compared $x^2$ with it by subtracting both
$x^2-x = x(x-1)$
so for $0<x<1$ this is always positive.
Then I tried to compare the first four terms of $\exp(x)$ with $x^2$ by dividing it by $x^2$ trying to find where it is greater than one
$\exp(x)-1 > x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$
$\frac{(x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!})}{x^2} = \frac{1}{x} + \frac{1}{2} + \frac{x}{6} + \frac{x^2}{24}$
I ignored the first term $\frac{1}{x}$ because if the rest of the equation was already bigger than $1$ for $x>1$ then the problem would be solved.
Solving the equation:
$\frac{1}{2} + \frac{x}{6} + \frac{x^2}{24} > 1$ with $x>0$
$\frac{x^2}{24} + \frac{x}{6} - \frac{1}{2}> 0$ with $x>0$
Its roots are at $x=-6$ and $x=2$. So for $x>2$ the equation is also valid
I still do not know how to prove that $\exp(x)-1>x^2$ for $1<x<2$.
Edit: the solution from rogerl is much better than mine, but just for completeness using the comment from JetfiRex
considering the first two terms:
$x + \frac{x^2}{2!} > x^2$
$x - \frac{x^2}{2} > 0$
$x(1 - \frac{x}{2}) > 0$
x=0 and x=2 are solutions of this downward parabola, so the equation is also valid from 0 to 2
 [/INST] $x+\frac{x^2}{2} + \frac{x^3}{6} > x^2$ reduces to
$$\frac{1}{6}x(x^2-3x+6) > 0,$$
which is true for all $x>0$ since the discriminant of the quadratic is negative.
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