Prove that $e^x\ge x+1$ for all real $x$ Without using differentiation, logarithmic function, rigorously, prove that $$e^x\ge x+1$$ for all real values of $x$.
 A: I don't know if it's cheating, but you didn't say that integration is forbidden. Since $\exp$ is increasing we know that $e^{-x} \leq 1$ for all $x \geq 0$. Integrating, we get
$$
\forall x\geq 0,\qquad 0 = \int_0^x 0\,dt \leq \int_0^x (1-e^{-t})\,dt = x + e^{-x} - 1.
$$
The inequality $\forall x \geq 0, \;e^x \geq 1 + x$ follows in the same lines as the former.
A: Write out the expansion of $(1 + {x \over n})^n$ as
$$(1 + {x \over n})^n = 1 + n {x \over n} + {n(n-1) \over 2}{x^2 \over n^2} + ...$$
If $x \geq 0$, then all terms are nonnegative, so the sum is at least the sum of the first two terms, namely $1 + x$. If $x \leq - 1$, then for even $n$, the expression $(1 + {x \over n})^n $ is nonnegative, so the limit must be at least zero, which is greater than or equal to $1 + x$.
So it remains to look at the case where $-1 < x < 0$. In this case write $x = -y$ and you have
$$(1 - {y \over n})^n = 1 - n {y \over n} + {n(n-1) \over 2}{y^2 \over n^2} - ...$$
This is an (finite) alternating series, and the absolute value of the ratio of two consecutive terms of this series is of the form ${n - k \over k+ 1} {y \over n}$, which is less than $1$ since $0 < y < 1$. So the terms decrease in absolute value. Hence the overall sum is at least what you get if you truncate after a negative term. So truncating after two terms you get
$$(1 - {y \over n})^n \geq 1 - y$$
Equivalently, 
$$(1 + {x \over n})^n \geq 1 + x$$
Taking limits as $n$ goes to infinity gives what you're looking for.
A: If $ x\ge 0 $ as $ \exp $ is increase we have $ e^{x} \ge 1 $. Then by mean value theorem there exist $ c \in (0,x) $ such that
\begin{equation}
e^x - 1 = e^c x.
\end{equation}
Thus
\begin{equation}
e^x \ge x +1.
\end{equation}
If $ x \le -1 $, we have 
\begin{equation}
e^x \ge 0 > 1+x
\end{equation}
Now if $ -1 < x < 0 $ we have 
\begin{equation}
e^c < 1 \Rightarrow -e^c > -1 \Rightarrow -e^c x < -x
\end{equation}
Thus
\begin{equation}
1 - e^x = e^c (-x) < -x
\end{equation}
and 
\begin{equation}
e^x  > 1+x 
\end{equation}
A: Bernoulli's Inequality: for any $\,n\in\Bbb N\,$
$$1+x\leq\left(1+\frac{x}{n}\right)^n\xrightarrow [n\to\infty]{} e^x$$
The inequality above is true for $\,x\geq -1\,$ , and since the wanted inequality is trivial for $\,x<-1\,$ we're done.
A: Hint:
Prove for rational $x=\frac{a}{b}$ ($b > 0$) that $e^{\frac{a}{b}} \ge \frac{a}{b}+1$.  You can do this by showing that
$$e^a \ge \left(\frac{a}{b}+1\right)^b.$$
Finally, argue by continuity.
A: Assuming $x > 0$ and regardless of your definition of $e$, you can show that $e^x = \displaystyle\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$.  Then:
\begin{align}
\left( 1 + \frac{x}{n} \right)^n &= \sum_{k=0}^n {n \choose k}\frac{x^k}{n^k}
\\
&= 1 + x + \sum_{k=2}^n {n \choose k} \frac{x^k}{n^k}
\end{align}
Since each term ${n \choose k} (x/n)^k$ is positive, it follows that
$$
\left( 1 + \frac{x}{n} \right)^n - x - 1 > 0.
$$
Hence,
$$
\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n - x - 1 \geq 0.
$$
from which the desired conclusion follows for $x > 0$.
A: Suppose $x \le 0$.
From $t^n - 1 = (t-1)(1 + t + \ldots + t^{n-1})$ (with $t = 1 + x/n$) we get
$$(1+x/n)^n - 1 = \frac{x}{n} \sum_{j=0}^{n-1} (1+x/n)^j$$
If $n$ is large enough that $1+x/n \ge 0$ we have $(1+x/n)^j \le 1$, so 
$$\sum_{j=0}^{n-1} (1+x/n)^j \le \sum_{j=0}^{n-1} 1 = n $$
and since $x/n \le 0$
$$ \frac{x}{n} \sum_{j=0}^{n-1} (1+x/n)^j \ge \frac{x}{n} n = x $$
Thus $$(1+x/n)^n \ge 1 + x$$
Now take the limit as $n \to \infty$.
A: Hint: Break it into two cases, $x \leq 0$ and $x > 0$. One of them is trivial. For the other case, consider the Taylor Series expansion.
