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

Question

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.

Answer 1

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 $y=1+x$.

Answer 2

Define $f(x)=e^x-x-1$. Notice that $f(0)=0$. As well, $f'(x)=e^x-1>0$ for all $x> 0$ and $f'(x)< 0$ for all $x<0$. It follows that $f(x)\geq 0 $ for all $x$ by mean value theorem.

Answer 3

The exponential function is concave up on the entire line. The line $y = 1 + x$ is tangent to its graph at $x = 0$. This line must lie below the graph of the exponential.

Answer 4

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 $x=0$.

Answer 5

If $x > 0$, the by the MVT for $f(t) = e^t-t-1$ on $(0,x)$: $$f(x)-f(0) = f'(r)(x-0) \Rightarrow\\ e^x-x-1 = (e^r-1)x > 0 \Rightarrow \\e^x-x-1 > 0,$$ and $x = 0$ LHS = RHS. If $x < 0 \Rightarrow$ apply the MVT on $(x,0)$ $$f(0) - f(x) = f'(r)(0-x)\Rightarrow \\0 -(e^x-x-1) = (e^r-1)(-x)<0\Rightarrow \\e^x-x-1 > 0.$$

Answer 6

Two lines proof:

The exponential function is convex and differentiable. Hence its graph lies above any of its tangent lines. Observe $y=1+x$ is the equation of the tangent line at $(0,1)$.