Is there a name for the "famous" inequality $1+x \leq e^x$? Is there a name for the "famous" inequality $1+x \leq e^x$? It has many variants depending on how you arrange the terms:

$$1 + x \leq e^x$$
$$e^{-x} -x - 1 \geq 0 $$
$$\ln(1+x) \leq x$$

Et cetera.

Perhaps the simplest mnemonic device is, "$e^x$ lies above its tangent line at the origin." This is at least a geometric instead of arbitrary algebraic expression of the fact.

It comes up in computer science and probability proofs quite frequently. In particular it is a lemma to Chernoff's bounds, and some results regarding the perceptron algorithm and Occam's razor in the PAC learning model. It is very easy to prove by drawing a graph or taking a derivative. 

Does it have a name?

More generally I want to ask "why" it's so important, but this is an extremely soft question and I only hope to get used to it in time.
It is so important
because it provides
a nice lower bound
for $e^x$
for 
small $x$.

The inequality
$\log(1+x) \ge 1+x$
for
$0 \le x$
follows easily from
$\log(x)
=\int_1^x \frac{dt}{t}
$.

It is a very special case
of the fact that
the tangent to a curve
at a point
is the best local linear
approximation
to the curve
at that point.

Analytically,
of course,
this is just
the first two terms in the
Taylor series:
$f(x+h)
\approx f(x)+hf'(x)+h^2f''(x)/2+...
$.
With the appropriate
remainder term,
this shows that,
if
$f''(x) \ge 0$,
then
$f(x+h)
\ge f(x)+hf'(x)
$.
The geometric intuition is certainly that $1+x$ is the tangent line to $e^x$ at $0$, so this amounts to the claim that "$e^x$ lies above its tangent line at the origin." The inequality is tight as $x \rightarrow 0$ in the sense that $e^x = 1 + x + o(x)$.

It's also the limit of the limit of Bernoulli $(1+x/n)^n \geq 1+x$, taking $n \rightarrow \infty$.

(Best answers were in question comments, so assembled here and self-accepted.)