The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is accessible to students with limited calculus experience, and does not involve too much analysis of different cases.
 89,705
 18
 102
 171
 1,177
 2
 8
 8

2Question about the same inequality: [Prove that $e^x\ge x+1$ for all real $x$](http://math.stackexchange.com/questions/252541/provethatexgex1forallrealx) – Martin Sleziak Sep 25 '13 at 12:31

1There are many great answers below. I now wonder whether, as a question with no objective right answer, this should be community wiki? – Ashley Montanaro Sep 26 '13 at 21:47

This is a duplicate of this : https://math.stackexchange.com/questions/252541/provethatexgex1forallrealx?noredirect=1&lq=1 – Guy Fsone Nov 10 '17 at 18:30
27 Answers
Another way (not sure if its "simple" though!): $y = x+1$ is the tangent line to $y = e^x$ when $x= 0$. Since $e^x$ is convex, it always remains above its tangent lines.
 42,360
 6
 33
 66

13That's my favourite argument. I'm afraid it fails on the "is accessible to students with limited calculus experience" criterion, but it's boootiful. – Daniel Fischer Sep 25 '13 at 12:28

18@DanielFischer: Are you sure that it fails? Tangent lines can appear much earlier then the derivatives and it have a nice graphical representation (sure, representation might not be mathematically precise). – Maciej Piechotka Sep 25 '13 at 13:38

4@Maciej I'm not sure. But without some calculus to build on, I think it would be too handwavy for me. – Daniel Fischer Sep 25 '13 at 13:44

19The convexity, at least, is elementary: $e^{x+z}  2 e^x + e^{xz} = (e^z1)^2 e^{xz} > 0$ for all real $x,z$, with equality **iff** $z=0$. – Noam D. Elkies Sep 25 '13 at 14:55

25

@DanielFischer while I agree with you to some extend I'm afraid that the handwavyness often comes in hindsight  you know it is handwavy because you know calculus and have experience with it. – Maciej Piechotka Sep 26 '13 at 20:44

3@NoamD.Elkies How does that imply convexity? Is it midpoint convexity, which is equivalent as exp is continuous? – snulty Aug 19 '15 at 19:21

$$ e^x = \lim_{n\to\infty}\left(1+\frac xn\right)^n\ge1+x $$
by Bernoulli's inequality.
 23,385
 4
 58
 99
The shortest proof I could think of: $$1 + x \leq 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = e^x.$$
However, it is not completely obvious for negative $x$.
Using derivatives:
Take $f(x) = e^x  1  x$. Then $f'(x) = e^x  1$ with $f'(x) = 0$ if and only if $x = 0$. But this is a minimum (global in this case) since $f''(0) = 1 > 0$ (the second derivative test). So $f(x) \geq 0$ for all real $x$, and the result follows.
Another fairly simple proof (but it uses Newton's generalization of the Binomial Theorem which is often covered in precalculus):
We proceed by contradiction. Suppose the inequality does not hold, i.e., $e^x < 1 + x$ for some $x$. Then $e^{kx} < (1 + x)^k$. Now set $x = 1/k$ so that \begin{align*} e &< \left( 1 + \frac{1}{k} \right)^k\\ &= 1 + \frac{k}{1}\left( \frac{1}{k} \right)^1 + \frac{k(k  1)}{1 \cdot 2}\left( \frac{1}{k} \right)^2 + \frac{k(k  1)(k  2)}{1 \cdot 2 \cdot 3}\left( \frac{1}{k} \right)^3 + \cdots\\ &< 1 + \frac{k}{1}\left( \frac{1}{k} \right)^1 + \frac{k^2}{1 \cdot 2}\left( \frac{1}{k} \right)^2 + \frac{k^3}{1 \cdot 2 \cdot 3}\left( \frac{1}{k} \right)^3 + \cdots\\ &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots\\ &= e, \end{align*} which is absurd. Therefore $1 + x \leq e^x$ for all real $x$.
By the way, this is where $$e = \lim_{k \to \infty}\left( 1 + \frac{1}{k} \right)^k$$ comes from because $$\lim_{k \to \infty}\frac{k(k  1)}{k^2} = \lim_{k \to \infty}\frac{k(k  1)(k  2)}{k^3} = \cdots = \lim_{k \to \infty}\frac{k(k  1)(k  2) \cdots (k  n)}{k^{n + 1}} = \cdots = 1.$$
 9,670
 2
 30
 52

19

4

1In your third proof : Why should such a $x$ be of the form $1/k$ with $k$ a positive integer ? – user37238 Sep 26 '13 at 15:20

1@user37238 Here, $k$ is not necessarily an integer because, technically, the proof uses the [Binomial Series](http://mathworld.wolfram.com/BinomialSeries.html). This explains why I wrote the binomial coefficients using the [falling factorial](http://mathworld.wolfram.com/FallingFactorial.html). – glebovg Sep 26 '13 at 17:58

I'm confused: you assume (for a contradiction) that $e^{x}<1+x$ for *some* real $x$, but then you claim that you can set $x=1/k$ without further justification. Would you not need to prove that $1/k$ is a valid choice of $x$ under such an assumption? – Will R Jul 28 '15 at 08:52
Let $f(x) = e^x(1+x)$, then $f^\prime(x) = e^x1$. Hence $f^\prime(x)=0$ iff $x=0$. Furthermore $f^{\prime\prime}(0) = e^0=1>0$, thus $f(0)=0$ must be the global minimum of $f$, proving your claim.
 7,179
 15
 20

1I like the idea behind this proof : just study a function and show that this function is always nonnegative. But I think that it could be a little simpler if you say that $f'\ge 0$ on $[0,+\infty($ (and consequently $f$ increases on this interval) and $f'\le 0$ on $)\infty,0]$ (and $f$ decreases on this interval) so $f(0)=0$ is a global minimum of $f$. – user37238 Sep 25 '13 at 14:59
One that's not been mentioned so far(?): knowing that $$ 0 < e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots $$ proves the inequality except for $1 < x < 0$. But in that region $$ e^x  (1+x) = \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots $$ is an alternating series whose terms decrease in absolute value and start out positive. Therefore it is positive by the usual argument: group the terms as $$ e^x  (1+x) = \left( \frac{x^2}{2} + \frac{x^3}{3!} \right) + \left( \frac{x^4}{4!} + \frac{x^5}{5!} \right) + \cdots $$ and observe that each combined term is positive, QED.
(This actually works for $3 < x < 0$, but you still want to use $e^x > 0$ to prove the inequality for very negative $x$.)
 25,109
 1
 61
 81
Repeatedly using $1 + x \le \left(1 + \frac{x}{2} \right)^2$, we have \begin{align} 1 + x \le \left(1 + \frac x 2\right)^2 \le \left(1 + \frac x 4\right)^4 \le \left(1 + \frac x 8\right)^8 \le \dots \le \left(1 + \frac x {2^k}\right)^{2^k}. \end{align} Taking the limit of $k \rightarrow \infty$ yields $$ 1 + x \le e^x. \qquad\qquad(1) $$
Another proof using the technique in this post. By the AMGM inequality, $$ \sqrt[n]{1 \times \cdots \times 1 \times (1 + x)} \le \frac{1 + \dots + 1 + (1 + x)}{n} =1 + \frac{x}{n}. $$ So, $$ 1+x \le \left(1 + \frac{x}{n} \right)^n. $$ Taking the limit of $n \rightarrow \infty$ yields (1).
If $x \ge 0$ then $$\begin{align} e^x = 1 + \int_0^x e^t\,\mathrm dt &= 1 + \int_0^x\left( 1 + \int_0^t e^u \,\mathrm du\right)\,\mathrm dt \\&= 1+x + \int_0^x \int_0^t e^u\,\mathrm du\,\mathrm dt \ge 1+x\end{align}$$
If $x \le 0$ then $$\begin{align}e^x = 1  \int_x^0 e^t\,\mathrm dt &= 1  \int_x^0\left( 1  \int_t^0 e^u\,\mathrm du\right)\,\mathrm dt \\ &= 1+x + \int_x^0 \int_t^0 e^u\,\mathrm du\,\mathrm dt \ge 1+x\end{align}$$
For completeness, using $\exp(x)=1+x+\frac{1}{2}x^2+\dots$, the inequality is trivial for $x\ge 0$. It is also trivial for $x<1$.
It remains to show the case $1<x<0$. Replacing $x$ by $x$, one need to show $1x < e^{x}$ for $0<x<1$, or $$1+x+\frac{1}{2}x^2+\dots=e^x <\frac{1}{1x}=1+x+x^2+\dots,$$ we are done.
 1
 3
 35
 77
 7,374
 17
 35


Yes, of course. From the post, it seemed that you were inferring this from the series expansion. – Mark Viola Nov 21 '16 at 03:24

Beautiful answers, but nobody used The Mean Value Theorem. Apply MVT on $[0,x] $ for $x>0$. There is some $c\in (0,x)$ such that:
\begin{align} \frac{e^xe^0}{x0} = e^c > 1 \end{align} So \begin{align} e^x>1+x \end{align} Something similar can be done for $x<0$. Finally note that we have equality when $x=0$. So we get the desired result: \begin{align} e^x\geq 1+x \end{align}
 8,438
 1
 11
 37
Completing glebovg's answer :
the inequality $1+x \le e^x$ clearly holds for $x \leq 1$,
suppose $x \geq 1$ :
the series $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ can be written (grouping the terms in pair) :
$$e^x = 1 + x + \sum_{k \geq 1} \left( \frac{x^{2k}}{(2k)!} + \frac{x^{2k+1}}{(2k+1)!} \right)$$
$$e^x = 1 + x + \sum_{k \geq 1} x^{2k}\left( \frac{1}{(2k)!} + \frac{x}{(2k+1)!} \right)$$
$$e^x = 1 + x + \sum_{k \geq 1} x^{2k}\left( \frac{2k + 1 + x}{(2k+1)!} \right)$$
under the assumption $x \geq 1$, the $\sum$ part is clearly a sum of positive numbers.
We want to prove that $1+x\le e^x$ for any $x\in\mathbb R$. Setting $x=\log(u)$, this is equivalent to proving:
$$ 1+\log(u)\le u $$ for any $u\in (0, \infty)$.
This is true because:
$$ 1+\log(u)=1+\int_1^u\frac1tdt\le1+\int_1^u1dt=1+u1=u $$
Some care is needed to establish that the inequality is true for both $u\ge1$ and $0<u\le1$. In the second case, we can see this more clearly by writing:
$$ 1+\int_1^u\frac1tdt=1+\int_u^1\frac1tdt\le1+\int_u^11dt=11+u=u $$
 23,385
 4
 58
 99
There is an amusing proof that I found yesterday that $e^x>x$ for every $x\in \mathbb{R}$.
It is obvious that $e^x>x$ if $x<0$ since the LHS is positive and the RHS is negative.
Suppose that for some $a\ge 0$, the inequality $e^a\le a$ holds.
Then $a\ge e^a\ge 1$ since $e^a\ge e^0$ because $a\ge 0$. But now we can see that $a\ge 1$ and again, $a\ge e^a\ge e^1$ and so $a\ge e$. We continue applying the same observation and conclude that $a\ge e^{^{e}}$ and so on, which means that $a$ is unbounded which is a contradiction.
 8,745
 4
 26
 42
For positive values of $ x $ We can use the following characterization of $e^x$ $$ e^x=\lim_{t\to \infty} \Big( 1+\frac{1}{t}\Big)^{tx},\quad t> 0,x\geq 0. $$ The Bernoulli's inequality states that $(1 + y)^r \geq 1 + ry$ for every $r \geq 1$ and every real number $y \geq −1$. Then for $y=\frac{1}{t}$ and $t>0$ such that $r=tx\geq 1$ we have \begin{align} e^x= &\lim_{t\to \infty} \Big( 1+\frac{1}{t}\Big)^{tx}\\ \geq &\lim_{t\to \infty} \Big(1+\frac{1}{t}(tx) \Big)\\ = & 1+x \end{align}
 13,301
 4
 43
 79

I just came across this answer 8 years later haha. However, I believe that $(1+y)^r\geq 1+ry$ does not hold for $0 < r < 1$. For example, $(1+3)^{0.5} = 2$ while $1+3\cdot 0.5 = 2.5$. – Gareth Ma Dec 06 '21 at 08:30

@GarethMa Yes, you are correct. The inequality is valid for $r \geq 1$. I will make the correction. – Elias Costa Dec 06 '21 at 12:08
One which uses $\exp(x) = \frac 1{\exp(x)} $
$$ 1 + x \underset{ \text{obvious}\\ \text{for $x>0$}}{\lt} 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + ... = {1 \over 1  x + {x^2 \over 2!}  { x^31 \over 3!} + ... } \tag 1 $$
Now we replace $+x$ by its negative counterparts and get similarily
$$ 1  x \underset{ \quad \text{for $x>0$}\\ \text{but not obvious}}{\lt} 1  x + {x^2 \over 2!}  {x^3 \over 3!} + ... = {1 \over 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + ... }\tag 2$$
But now the comparision with the fraction on the rhs becomes obvious if we look at the reciprocals. The reciprocal ${1\over 1x}=1+x+x^2+x^3+...$ is and we get
$$ {1 \over 1  x} = 1+x+x^2+... \underset{ \text{obvious}\\ \text{for $x>0$}}{\gt}
1 + x + {x^2 \over 2!} + {x^3 \over 3!} + ... = {1 \over 1  x + {x^2 \over 2!}  {x^3 \over 3!} + ... }\\ \tag 3 $$
 32,738
 3
 60
 134

1upps, I see there was another answer earlier, but which seemed too short for me to not only skim over it... – Gottfried Helms Sep 26 '13 at 09:28
For $x>0$ we have $e^t>1$ for $0<t<x$
Hence, $$x=\int_0^x1dt \color{red}{\le} \int_0^xe^tdt =e^x1 \implies 1+x\le e^x$$ For $x<0$ we have $e^{t} <1$ for $x <t<0$
$$x=\int^0_x1dt \color{red}{\ge} \int^0_xe^tdt =1e^x \implies 1+x\le e^x$$
 22,154
 4
 51
 94
Another simple proof...
Define function $f(x)=e^x(x+1)$. The minimum value is $0$ at $x=0$, it's also convex, so $f(x) \ge 0$.
 2,540
 11
 14
Proof by induction (works for natural numbers)
Assume it works for n
1 + n < e^n
Then we prove that it works for n+1
1 + (n+1) < e^(n+1)
Proof
1 + n < e^n
or 1 + n + 1 < e^n + 1
or 1 + n + 1 < e^n + e^n since e^n > 1
or 1 + n + 1 < e^n * 2
or 1 + (n+1) < e^n * e since e > 2
or 1 + (n+1) < e^(n+1)
hence it is true for n+1 if true for n. We know it is true for 1, hence by induction is true of 2, 3, 4...so on.
 147
 3


1

@tihom You can get a general proof from your proof in the following way: $floor(x)+1>x\geq floor(x) $ where $floor(x)$ is the greatest integer function. Now $$e^x\geq e^{floor(x)}> floor(x)+1> x$$ whenever $x\geq 1$ – Sedergine Sep 18 '21 at 02:07

More precisely, this was for $e^x>x$ but refining your inductive argument to $e^n>n+2$ for $n\geq 2$ we can easily fix it. – Sedergine Sep 18 '21 at 02:34
For $x > 0$, consider the mean value theorem on the interval $[0,x]$. Then $$e^x  e^0 \geq \inf_{(0,x)} e^c \cdot (x  0) = x,$$ implying $e^x \geq 1+x$. For $x < 0$, apply MTV on $[x,0]$: $$e^0  e^x \leq \sup_{(x,0)} e^c \cdot (0  x) = x,$$ giving us $e^x \leq x  1$, or $e^x \geq x + 1$. Then check $x = 0$ and the equality is proven.
 5,607
 1
 15
 29
A proof using only a little basic calculus, and not too many cases:
Set
$\alpha(x) = e^{x}(1 + x); \tag{1}$
then
$\alpha'(x) = e^{x}(1 + x) + e^{x} = xe^{x}, \tag{2}$
and
$\alpha(0) = 1; \tag{3}$
we note that
$x > 0 \Rightarrow \alpha'(x) < 0 \tag{4}$
and
$x < 0 \Rightarrow \alpha'(x) > 0 \tag{5}$
with
$\alpha'(0) = 0; \tag{6}$
thus, for $x > 0$,
$\alpha(x)  1 = \alpha(x)  \alpha(0) = \int_0^x \alpha'(s) ds < 0, \tag{7}$
whence
$e^{x}(1 + x) = \alpha(x) < 1; \tag{8}$
when $x < 0$,
$1  \alpha(x) = \int_x^0 \alpha'(s) ds > 0, \tag{9}$
yielding
$e^{x}(1 + x) = \alpha(x) < 1 \tag{10}$
in this case as well; (8) and (10) together imply
$1 + x < e^x \tag{11}$
when $x \ne 0$; clearly
$1 + 0 = 1 = e^0; \tag{12}$
combining (11) and (12) shows that
$1 + x \le e^x \tag{13}$
for every $x \in \Bbb R$, with strict inequality precisely when $x \ne 0$. QED.
 1
 3
 35
 77
 68,171
 5
 53
 112
We know the function $x^x$ has a single local minimum at $x=\frac1e$. Thus, for positive $x$, we have: \begin{align} \left(\frac1e\right)^{1/e}&\le x^x\\ e^{1/e}&\ge\frac1{x^x}\\ e^{1/xe}&\ge\frac1x\\ e^{1/xe1}&\ge\frac1{xe}\\ e^{(1/xe1)}&\ge\left(\frac1{xe}1\right)+1 \end{align} Let $t=\frac1{xe}1$. If $x>0$, we have $t>1$. Thus, for all $t>1$: $$e^t\ge t+1$$ (To prove the above for $t\le 1$, simply note that the lefthand side is always positive while the righthand side would be zero or negative.) QED.
 19,084
 1
 31
 84
The approximation of the exponential function by its linear Taylor polynomial has remainder term $R(x) := \exp(x)  (1+x)$. The Taylor Remainder Theorem then yields some $\xi$ in between $0$ and $x$ such that $R(x) = \exp''(\xi) x^2 = \exp(\xi) x^2 \ge 0$.
 119
 2
 11
Let $f(x)=\exp(x)x1$. Then, $f'(0)=0$. But $f$ is strictly convex (a difference of a strictly convex function and an affine one), so that $0$ most be a unique global minimum. Hence, $\exp(x)x1=f(x)\geq f(0)=0$ for all $x\in\mathbb{R}$.
 22,017
 3
 33
 79
We want to show that (1) $$1+x\leq e^x,$$ for $x\in\mathbb{R}$. When $x\geq 0$, we have $$1+x\leq 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=e^x.$$ Suppose $x=X$, where $X>1$, then $1+x=1X<0$ and $e^{x}=e^{X}=1/e^X>0$. Hence (1) holds.
Now take logarithms of (1) to obtain $$\log(1+x)\leq x.$$ But $$\log(1+x) = x\frac{x^2}{2}+\frac{x^3}{3}\cdots,$$ where $x<1$. Suppose $x=X$, where $0<X<1$, then $$\log(1+x)=\log(1X)=X\frac{X^2}{2}\frac{X^3}{3}\cdots<X=x.$$ Or, equivalently, $$1+x<e^x,$$ where $1<x<0$.
 9,831
 5
 40
 73

1Third line, I think you meant "Suppose $x=X$" and not "Suppose $x<X$". – user37238 Sep 25 '13 at 12:49
The fact $\frac{d}{dx} e^x = e^x$ is nicely demonstrated using the selfsimilar nature of exponential functions. (See my answer here.)
This justifies (actually, declares) that $y=x+1$ is tangent to $y=e^x$; thereafter, since the slope increases (or decreases) as $x$ gets larger (respectively, smaller) the line and curve cannot meet again (which is an informal way of stating the convexity property).
This answer uses no calculus or geometry.
Prerequisites: Algebra, basic facts about limits and that for $a \gt 0$ we can define $a^x$ for $x \in \Bbb R$
(see limits of rational exponents).
Theorem 1: There is one and only one number $a \gt 0$ satisfying
$$\tag 1 \forall x \in \Bbb R, \; a^x \ge 1 + x$$
Analyzing $\text{(1)}$, you'll be naturally lead to examine
$\tag 2 u_n \le a \le v_n \text{ where } n \ge 2$
with
$\tag 3 u_n = (1 + \frac{1}{n})^n \text{ and } \le v_n = (1  \frac{1}{n})^{n}$
Searching, you find answer links from this site:
$\quad u_n \text{ is strictly increasing}:\quad$ here
$\quad v_n \text{ is algebraically related to } u_n:\quad$ here
$\quad u_n \le 3:\quad$ here
By working with the theory in the above links you will conclude that only one real number, call it $e$, can possibly satisfy $\text{(1)}$. Again, as in the first answer link above, you will use the Bernoulli's inequality and
$\tag 4 e^\frac{s}{t} =\lim_{n\to \infty} \Big( 1+\frac{\frac{s}{t}}{\frac{ns}{t}} \Big)^{\frac{ns}{t}}$
to wrap things up:
$$ \forall x \in \Bbb R, \; e^x \ge 1 + x$$
 10,521
 1
 17
 43
The series expansion of $(1+x)$ is $(1+x)$, while $\exp(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+...$. Subtracting the second from the first one you have the difference $d=\exp(x)(1+x)=\frac{1}{2}x^2+\frac{1}{6}x^3+...$ which is zero only for $x=0$ otherwise $d\gt 0$ Q.E.D.
 10,172
 4
 35
 71

3

@Macavity: make the substitution: $x\to y$. Anyway it's not trivial. – Riccardo.Alestra Sep 25 '13 at 10:26

I have seen that proof done considering three regions separately  viz. $x \ge 0, 1 < x < 0, x \le 1$. The first region is trivial, the rest two not so much :( – Macavity Sep 25 '13 at 10:34
$1+x \le e^x$
Take the ln of both sides
$\ln(1+x) \le x$
differentiate both sides w.r.t $x$
$\frac{1}{1+x} \le 1$
which holds $\forall x \in \mathbb{R}, x \neq 1$.
 1
 3
 35
 77
 17
 1

3You should work this proof in the opposite direction for it to be an actual proof. – Cameron Williams Sep 25 '13 at 14:51

5$e^{x}<0$. Now differentiate both sides w.r.t. $x$ to get $e^{x}<0$, which holds $\forall x\in\mathbb R$. So our original statement is correct, is it? – John Gowers Sep 25 '13 at 14:54


3@GottfriedHelms  I just proved it! Just like the OP, I differentiated both sides with respect to $x$ and arrived at a manifestly true statement. So it must be true! – John Gowers Sep 26 '13 at 09:05

1@donkey: put any real number $x$ into your calculator and compute $\exp(x)$. The result will always be greater than zero, irrespectively of whether $x$ is positive or negative. So there must be a flaw in your proof... – Gottfried Helms Sep 26 '13 at 09:19

1@GottfriedHelms What is this? But we just proved that $e^{x}$ is always negative! A contradiction in mathematics!! (**Hint:** I was not being entirely serious with my comment.) – John Gowers Sep 26 '13 at 09:21

1@donkey: :). Perhaps a short coffeebreak should help to reconsider this later and find the error... – Gottfried Helms Sep 26 '13 at 09:26