How to prove $1+x \leq e^x~\forall x \in \mathbb{R}?$

How to prove $$1+x \leq e^x~\forall x \in \mathbb{R}$$

I'm stuck, I tried taking logs but didn't know how to proceed.

Solution 1:

Proof 1: Bernoulli's inequality

We use Bernoulli's inequality: $$(1+u)^n\geq 1+nu;\ u\geq -1, \text{ integer } n\geq 0$$

Then for $n>|x|$, we have that $$\left(1+\frac{x}{n}\right)^n \geq 1+x$$

Then take the limit: $$e^x = \lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n \geq 1+x$$

Proof 2: Mean Value Theorem

Let $x>0$ first. By the mean value theorem, there is an $c\in(0,x)$ such that:

$$\frac{e^x-e^{0}}{x-0}=e^{c}.$$ Since $c> 0$, $e^c> 1$, so we get $e^x> 1+x$.

If $x<0$ then for some $c\in(x,0),$ $$\frac{e^0-e^x}{0-x} =e^{c}.$$ Since $e^{c}<1$ we conclude:

$$1-e^x< -x\implies 1+x< e^x$$

Finally, just check $x=0$.

Solution 2:

There are lots of ways to prove this inequality.

For one: if you know Taylor series, then you know that $$ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots\geq1+x. $$ (This inequality is clearly true for $x\geq 0$; it is also clearly true for $x\leq -1$, since $e^x>0$. It requires some argument for other values of $x$; try considering pairing up subsequent terms in the series.)

You can also use convexity: you know that $\frac{d^2}{dx^2}[e^x]=e^x>0$ for all $x$; so, $f(x)=e^x$ is a convex function. As such, it lies above all of its tangent lines. The tangent line to $y=e^x$ at $x=0$ is $y=1+x$.