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.
 A: 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 $x=0$.
A: 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 $y=1+x$.
A: 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 $(0,1)$.
A: 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. 
A: 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.
A: 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 $(x,0)$ $$f(0) - f(x) = f'(r)(0-x)\Rightarrow \\0 -(e^x-x-1) = (e^r-1)(-x)<0\Rightarrow \\e^x-x-1 > 0.$$
