Proof that $e^{-x} \ge 1-x$

Solution 1:

Recall the Mean Value Theorem that says that for a continuous function on $[a,b]$, and a differentiable function on the interval $(a, b)$, there exists a $c \in (a,b)$ such that

$$\frac{f(b)-f(a)}{b-a} = f^{\prime}(c).$$

Consider working on the interval $[0, x].$ Then, in our case,

$$\frac{e^{-x} - 1}{x} = -e^{-c}$$

for some $c \in (0, x)$. It is clear to see that for any $c$ lying in that interval, $-e^{-c} \geq -1.$ Hence,

$$\frac{e^{-x}-1}{x} \geq -1.$$

The result then follows,

$$e^{-x}-1 \geq -x \implies e^{-x} \geq 1 -x$$

for any $x \geq 0.$

Solution 2:

Consider the function and its derivatives $$f(x)=e^{-x}-1+x\qquad f'(x)=1-e^{-x}\qquad f''(x)=e^{-x}$$ The first derivative cancels for $x=0$ and $f(0)=0$. The second derivative is always positive; so $x=0$ corresponds to the minimum and f(x) is always greater or equal $0$. So, the relation holds for any $x$ (positive or negative; the sign does not matter).

Solution 3:

CLAIM

Here, I provide a alternative proof of $$e^x \ge x+1$$ for all real $x$.

Let $f(x) = e^x-(1+x)$, then $f^\prime(x) = e^x-1$.

Hence $f^\prime(x)=0$ iff $x=0$.

Furthermore, note that $f^{\prime\prime}(x) = e^x>0$.

Thus, $f(0)=0$ must be the global minimum of $f(x)$.

Our inequality is proven.

HOW TO APPLY

Let $x=-t$. Then $e^{-t} \ge -t+1=1-t$ for $t$.

In fact, this goes further-it proves that it is true not only for $t \ge 0$, but for any real $t$.