Prove that $e^x\ge x+1$ for all real $x$

Question

Without using differentiation, logarithmic function, rigorously, prove that $$e^x\ge x+1$$ for all real values of $x$.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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}

Answer 5

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.

Answer 6

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 $x > 0$.

Answer 7

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 $n \to \infty$.

Answer 8

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.