TITLE: If an inequality is true for all natural numbers, is it necessarily true for all real numbers inbetween?
QUESTION [12 upvotes]: A lot of the time in lectures, my professors prove (by induction) an inequality (e.g. $(1+x)^n \geq 1+nx$) in the natural numbers (or any subsets thereof), and I've noticed (not rigourously; only by graphing the functions) that such statements are also true for all real numbers inbetween.
Another example is that exponential growth beats polynomial growth.
My question is:

If an inequality is true for all $n \in \mathbb{N},$ does it necessarily follow that the same inequality is true for all $n \in \mathbb{R^+}$?

I'm not in the market for a rigourous proof; (if the answer's no) just  a counter-example, or (if the answer's yes) an intuitive reason why this is the case.

REPLY [3 votes]: Without further requirements on the inequality, the answer is a massive no, as all respondent say. This is because there is not constraint at all between neighboring values of the argument.
The answer would be different and much more interesting if you imposed smoothness conditions, like continuity to some order, differentiability to some order, Lipschitz continuity, band-limitedness...
Your question is probably hiding a deeper one: given a discrete function, is there a "natural" way to define an extension to real values ? And what properties would it possess ?
A good example of such an extension is given by the Gamma function, $\Gamma(x+1)$, that generalizes the factorial $n!$.
If your inequality is $n!\ge1$, it is violated between $0!$ and $1!$, but this is quite understandable as the two first values form a plateau.
http://en.wikipedia.org/wiki/File:Factorial_Interpolation.svg