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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.

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27 Answers 27

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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.

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    $\begingroup$ That's my favourite argument. I'm afraid it fails on the "is accessible to students with limited calculus experience" criterion, but it's boootiful. $\endgroup$ Sep 25, 2013 at 12:28
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    $\begingroup$ @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). $\endgroup$ Sep 25, 2013 at 13:38
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    $\begingroup$ @Maciej I'm not sure. But without some calculus to build on, I think it would be too hand-wavy for me. $\endgroup$ Sep 25, 2013 at 13:44
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    $\begingroup$ The convexity, at least, is elementary: $e^{x+z} - 2 e^x + e^{x-z} = (e^z-1)^2 e^{x-z} > 0$ for all real $x,z$, with equality iff $z=0$. $\endgroup$ Sep 25, 2013 at 14:55
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    $\begingroup$ @NoamD.Elkies -- for large values of "elementary". $\endgroup$ Sep 25, 2013 at 18:52
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$$ e^x = \lim_{n\to\infty}\left(1+\frac xn\right)^n\ge1+x $$

by Bernoulli's inequality.

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    $\begingroup$ This is what I thought of when I saw this question. (+1) $\endgroup$
    – robjohn
    Sep 27, 2013 at 0:31
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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 $$\lim_{k \to \infty}\frac{k(k - 1)}{k^2} = \lim_{k \to \infty}\frac{k(k - 1)(k - 2)}{k^3} = \cdots = \lim_{k \to \infty}\frac{k(k - 1)(k - 2) \cdots (k - n)}{k^{n + 1}} = \cdots = 1.$$

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    $\begingroup$ Is it that evident for negative $x$? $\endgroup$
    – Macavity
    Sep 25, 2013 at 10:16
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    $\begingroup$ @Macavity for x<-1 is obvious $\endgroup$ Sep 25, 2013 at 17:35
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    $\begingroup$ In your third proof : Why should such a $x$ be of the form $1/k$ with $k$ a positive integer ? $\endgroup$
    – user37238
    Sep 26, 2013 at 15:20
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    $\begingroup$ @user37238 Here, $k$ is not necessarily an integer because, technically, the proof uses the Binomial Series. This explains why I wrote the binomial coefficients using the falling factorial. $\endgroup$
    – glebovg
    Sep 26, 2013 at 17:58
  • $\begingroup$ 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? $\endgroup$
    – Will R
    Jul 28, 2015 at 8:52
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Let $f(x) = e^x-(1+x)$, then $f^\prime(x) = e^x-1$. 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.

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    $\begingroup$ I like the idea behind this proof : just study a function and show that this function is always non-negative. 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 $f$. $\endgroup$
    – user37238
    Sep 25, 2013 at 14:59
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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 $x$.)

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    $\begingroup$ Why does knowing that $0<e^x$ give the result for $x\le-1$? $\endgroup$ Sep 25, 2013 at 14:56
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    $\begingroup$ $x \leq -1$ means $x+1 \leq 0 < e^x$. $\endgroup$ Sep 25, 2013 at 14:58
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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 $$\begin{align}e^x = 1 - \int_x^0 e^t\,\mathrm dt &= 1 - \int_x^0\left( 1 - \int_t^0 e^u\,\mathrm du\right)\,\mathrm dt \\ &= 1+x + \int_x^0 \int_t^0 e^u\,\mathrm du\,\mathrm dt \ge 1+x\end{align}$$

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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 AM-GM 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).

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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 $1-x < e^{-x}$ for $0<x<1$, or $$1+x+\frac{1}{2}x^2+\dots=e^x <\frac{1}{1-x}=1+x+x^2+\dots,$$ we are done.

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  • $\begingroup$ @Dr.MV $x+1<0< e^x$ $\endgroup$
    – Ma Ming
    Nov 21, 2016 at 2:47
  • $\begingroup$ Yes, of course. From the post, it seemed that you were inferring this from the series expansion. $\endgroup$
    – Mark Viola
    Nov 21, 2016 at 3:24
  • $\begingroup$ Did you mean it's trivial for $x \mathbf \le -1$ ? $\endgroup$ Mar 14, 2019 at 0:31
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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.

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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^x-e^0}{x-0} = 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}

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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:

$$ 1+\log(u)=1+\int_1^u\frac1tdt\le1+\int_1^u1dt=1+u-1=u $$

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:

$$ 1+\int_1^u\frac1tdt=1+\int_u^1-\frac1tdt\le1+\int_u^1-1dt=1-1+u=u $$

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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.

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    $\begingroup$ It's actually a great proof $\endgroup$
    – explogx
    Jan 17, 2021 at 0:24
  • $\begingroup$ @coshsinh Thank you. $\endgroup$ Jan 17, 2021 at 9:09
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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}

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  • $\begingroup$ 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 $1+3\cdot 0.5 = 2.5$. $\endgroup$
    – Gareth Ma
    Dec 6, 2021 at 8:30
  • $\begingroup$ @GarethMa Yes, you are correct. The inequality is valid for $r \geq 1$. I will make the correction. $\endgroup$ Dec 6, 2021 at 12:08
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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 ${1\over 1-x}=1+x+x^2+x^3+...$ is and we get $$ {1 \over 1 - x} = 1+x+x^2+... \underset{ \text{obvious}\\ \text{for $x>0$}}{\gt} 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + ... = {1 \over 1 - x + {x^2 \over 2!} - {x^3 \over 3!} + ... }\\ \tag 3 $$

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    $\begingroup$ upps, I see there was another answer earlier, but which seemed too short for me to not only skim over it... $\endgroup$ Sep 26, 2013 at 9:28
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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.

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  • $\begingroup$ This isn't really a proof. Like, where is your inductive step? $\endgroup$
    – Alexander
    Sep 25, 2013 at 23:00
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    $\begingroup$ updated the answer works for natural number only though $\endgroup$
    – tihom
    Sep 25, 2013 at 23:20
  • $\begingroup$ @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 $x\geq 1$ $\endgroup$
    – Sedergine
    Sep 18, 2021 at 2:07
  • $\begingroup$ 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. $\endgroup$
    – Sedergine
    Sep 18, 2021 at 2:34
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For $x>0$ we have $e^t>1$ for $0<t<x$

Hence, $$x=\int_0^x1dt \color{red}{\le} \int_0^xe^tdt =e^x-1 \implies 1+x\le e^x$$ For $x<0$ we have $e^{t} <1$ for $x <t<0$

$$-x=\int^0_x1dt \color{red}{\ge} \int^0_xe^tdt =1-e^x \implies 1+x\le e^x$$

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Another simple proof...

Define function $f(x)=e^x-(x+1)$. The minimum value is $0$ at $x=0$, it's also convex, so $f(x) \ge 0$.

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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.

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Let $f(x)=\exp(x)-x-1$. 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, $\exp(x)-x-1=f(x)\geq f(0)=0$ for all $x\in\mathbb{R}$.

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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 $1+x=1-X<0$ 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 $$\log(1+x)=\log(1-X)=-X-\frac{X^2}{2}-\frac{X^3}{3}-\cdots<-X=x.$$ Or, equivalently, $$1+x<e^x,$$ where $-1<x<0$.

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    $\begingroup$ Third line, I think you meant "Suppose $x=-X$" and not "Suppose $x<-X$". $\endgroup$
    – user37238
    Sep 25, 2013 at 12:49
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The fact $\frac{d}{dx} e^x = e^x$ is nicely demonstrated using the self-similar 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).

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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/xe-1}&\ge\frac1{xe}\\ e^{(1/xe-1)}&\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 left-hand side is always positive while the right-hand side would be zero or negative.) QED.

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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.

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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 $R(x) = \exp''(\xi) x^2 = \exp(\xi) x^2 \ge 0$.

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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.

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    $\begingroup$ Same comment, is it trivially evident for negative $x$ that $d>0$? $\endgroup$
    – Macavity
    Sep 25, 2013 at 10:17
  • $\begingroup$ @Macavity: make the substitution: $x\to -y$. Anyway it's not trivial. $\endgroup$ Sep 25, 2013 at 10:26
  • $\begingroup$ 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 :( $\endgroup$
    – Macavity
    Sep 25, 2013 at 10:34
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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:

$$ \forall x \in \Bbb R, \; e^x \ge 1 + x$$

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$1+x \le e^x$

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 $\forall x \in \mathbb{R}, x \neq -1$.

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    $\begingroup$ You should work this proof in the opposite direction for it to be an actual proof. $\endgroup$ Sep 25, 2013 at 14:51
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    $\begingroup$ $e^{-x}<0$. 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? $\endgroup$ Sep 25, 2013 at 14:54
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    $\begingroup$ @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! $\endgroup$ Sep 26, 2013 at 9:05
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    $\begingroup$ @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... $\endgroup$ Sep 26, 2013 at 9:19
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    $\begingroup$ @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.) $\endgroup$ Sep 26, 2013 at 9:21

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