By expanding $e^x$ into a series prove the following inequality By expanding $e^x$ into a series $\sum e^x$ prove that $$\forall x \in \mathbb{R}, x \ge 0 \implies e^{x-1} \ge x$$
Also show when this inequality becomes equality.
I'm not really sure how to attack this problem at all, any tips would be very welcome.
 A: Series expansion is not really needed. Since $e^x$ is a convex function, its graphics lies above any tangent line, so, by considering the tangent line in $x=0$, we have $e^x\geq x+1$ for any $x\in\mathbb{R}$, with equality only in $x=0$, or
$$ e^{x-1}\geq x $$
with equality only in $x=1$, as wanted (just replace $x$ with $x-1$).
A: Set
$y = x - 1; \tag{1}$
then
$x = y + 1, \tag{2}$
so that the inequality
$e^{x - 1} \ge x \tag{3}$
holds if and only if
$e^y \ge 1 + y. \tag{4}$
The inequality (4) is demonstrated in my answer to this question; then (3) immediately follows.  Strict inequality when $y \ne 0$ is also therein addressed.  QED.
Notes:
(1.)  So how does the power series
$e^x = \sum_0^\infty \dfrac {x^n}{n!} \tag{5}$
enter into this discussion?  Well, inspection of the proof presented in the the linked citing shows that it rests solidly upon the fact that
$(e^x)' = \dfrac{de^x}{dx} = e^x; \tag{6}$
but (5) readily implies (6), via term-by-term differentiation; indeed, we see from (5) that
$(e^x)' = \sum_0^\infty (\dfrac{x^n}{n!})' = \sum_0^\infty \dfrac{nx^{n -1}}{n!} = \sum_1^\infty \dfrac{x^{n -1}}{(n - 1)!} = \sum_0^\infty \dfrac {x^n}{n!} = e^x.  \tag{7}$
Then the only issue which remains is verification that term-by-term differentiation is valid for the series (5); but this follows since the exponential series (5) is both absolutely and uniformly convergent on any closed interval $[-M, M]$ for $) \le M \in \Bbb R$.  Details may be found here in gower's blog.
(2.)  It may be asked if there isn't a proof of (3)-(4) which more directly uses the power series (5).  In response, I would like to say that any demonstration of (3)-(4) from (5) is certain to require some kind of fiddling around with the power series, especially for $y < 0$ (the result is trivial for positive $y$).  So I figure my little fiddle tune is as good as almost any, even if it doesn't qualify for first violin.  But then again, maybe it does! ;-) End of Notes.
