TITLE: How do I show that this function is always $> 0$
QUESTION [6 upvotes]: Show that $$f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +
 \frac{x^4}{4!} > 0 ~~~ \forall_x \in \mathbb{R}$$

I can show that the first 3 terms are $> 0$ for all $x$:
$(x+1)^2 + 1 > 0$
But, I'm having trouble with the last two terms. I tried to show that the following was true:
$\frac{x^3}{3!} \leq \frac{x^4}{4!}$
$4x^3 \leq x^4$
$4 \leq x$
which is not true for all $x$.
I tried taking the derivative and all that I could ascertain was that the the function became more and more increasing as $x \rightarrow \infty$ and became more and more decreasing as $x \rightarrow -\infty$, but I couldn't seem to prove that there were no roots to go with this property.

REPLY [12 votes]: Hint:
$$f(x) = \frac{1}{4} + \frac{(x + 3/2)^2}{3} +\frac{x^2(x+2)^2}{24}$$

REPLY [3 votes]: $f$ is a polynomial, and therefore, is differentiable at all points.  Furthermore, as $x\to\infty$ or $x\to-\infty$, $f(x)\to+\infty$.  Thus, if $f(x)\le0$ for some $x$, then $f(x)\le0$ for some relative minimum.
$$f'(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$$
$f'(x)=0$ for all relative minima.  However, if $f'(x)=0$, then $$f(x)=f'(x)+\frac{x^4}{4!}$$
Thus, $f(x)=\frac{x^4}{4!}>0$ for all relative minima $x\not=0$.  $x=0$ is not a relative minimum, because $f'(0)\not=0$, so this equation holds for all relative minima of $f$.  This contradicts our assumption, so $f(x)>0$ for all $x\in \mathbb R$.