How do you prove $e^{x} \geq \left(1+\frac{x}{n}\right)^{n}$ for $n \geq 1$ How do you prove $e^{x} \geq \left(1+\frac{x}{n}\right)^{n}$ for $n \geq 1$?
I can prove this for natural numbers only via induction, but how do you prove this for any real $n \geq 1$?
We start with the base case $n=1$. We have $e^x \geq 1+x$ by a variety of methods.
For the induction step, assume $e^{x} \geq \left(1+\frac{x}{n}\right)^{n}$.
Notice that taking the derivative of $(1+\frac{x}{n+1})^{n+1}$ gives us $(1+\frac{x}{n+1})^{n}$ and thus $(1+\frac{x}{n+1})^{n} < \left(1+\frac{x}{n}\right)^{n} \leq e^x = \frac{d}{dx} e^x$.
I'm not sure how to extend this to the non-integer case. Any help would be appreciated.
 A: In this answer I define the natural logarithm as
$$
\log x =\int_{1}^{x}\frac{1}{t} \, dt \, .
$$
and the exponential function as the inverse of the logarithm. This leads to the definition $a^x=\exp(x\log a)$ for $a>0$, valid for all real $x$.

We begin with the fact that
$$
\frac{x}{n} \geq \log\left(1+\frac{x}{n}\right) \, .
$$
which can be deduced geometrically:

$\log\left(1+\frac{x}{n}\right)$ is the area of the region bounded by the hyperbola $y=1/t$, the $t$-axis, and the vertical lines $t=1$ and $t=1+\frac{x}{n}$. The rectangle with a width of $x/n$ and a height of $1$ gives us an upper bound for this region. A little care is needed when $-1 < x/n < 0$, but the result still holds. Hence,
$$
n\log\left(1+\frac{x}{n}\right) \leq x \, ,
$$
and therefore
$$
\exp\left(n\log\left(1+\frac{x}{n}\right)\right)=\left(1+\frac{x}{n}\right)^n \leq \exp(x) \, .
$$
Note however that this proof is only valid for $\frac{x}{n}>-1$. If $\frac{x}{n} \le -1$, then $(1+\frac{x}{n})$ is nonpositive, and so the definition $\exp\left(n\log\left(1+\frac{x}{n}\right)\right)=\left(1+\frac{x}{n}\right)^n$ no longer applies.
A: Work with the logarithm of both terms ...
Set $h(x) = x - n\log(1 + x/n)$ on $0 \leq x < \infty$. You have $h(0) = 0$ and
\begin{align}
h'(x) &= 1 - 1/(1 + x/n) \\
&= \frac{x/n}{1 + x/n},
\end{align}
which is nonnegative, so $h$ is increasing for $x > 0$, that is, $h(x) \geq 0$.
