Short proof for $1-x\leq \exp(-x)$

Question

I'm trying to find a short way to show that: $$ 1-x\leq \exp(-x),\forall x\in\mathbb{R} $$

Does anyone know a really short solution for this problem?

Set $y=-x$ and apply your favourite proof from http://math.stackexchange.com/questions/504663/simplest-or-nicest-proof-that-1x-le-ex/504837 — John Gowers, Oct 08 '15 at 20:17

score 4 · Accepted Answer · answered Oct 08 '15 at 20:12

4

Hint: try to minimize the function $$ f(x) = \exp(-x) - 1 + x $$ using calculus. You will find that its unique minimum occurs at $x = 0$.

answered Oct 08 '15 at 20:12

Ben Grossmann

score 4 · Answer 2 · answered Oct 08 '15 at 20:14

4

$\exp(-x)$ is convex, so it's above its tangents, especially its tangent at 0.

answered Oct 08 '15 at 20:14

Augustin

score 3 · Answer 3 · edited Oct 08 '15 at 20:23

3

Another approach is to recall that

$$\left(1-\frac xn\right)^n$$

is a inceasing function of $n$ and is bounded above by its limit, $e^{-x}$. Therefore

$$e^{-x}\ge \left(1-\frac xn\right)^n\ge 1-x$$

and we're done.

edited Oct 08 '15 at 20:23

Travis Willse

answered Oct 08 '15 at 20:20

Mark Viola

A nice alternative – Simon S Oct 08 '15 at 20:26
@travis Just curious... am at the gym and working from a "not so smart" phone. What was the edit? – Mark Viola Oct 08 '15 at 20:30
@simons Thank you! Much appreciative. – Mark Viola Oct 08 '15 at 20:30
@Dr.MV There was a caret missing from one of the equations, so that $e^{-x}$ rendered as $e - x$. (+1 for the nice answer, btw.) – Travis Willse Oct 08 '15 at 20:31
@Travis Thanks for the edit and the +1!! – Mark Viola Oct 08 '15 at 20:42

score 1 · Answer 4 · edited Apr 13 '17 at 12:20

In my answer to this question, Simplest or nicest proof that $1+x \le e^x$, as well as in my answer to this one: To show that $e^x > 1+x$ for any $x\ne 0$, I proved that

$1 + x \le e^x \tag{1}$

for all real $x$. Many others also contributed concise and valid proofs of this fact as well; they deserve credit a much as anyone.

Given that (1) is the case, it follows that

$1 - x \le e^{-x} \tag{2}$

for all $x \in \Bbb R$ by simply replacing $x$ by $-x$ throughout. QED.

4 Answers4