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My aim is to prove that $e^{-x} \geq 1-x$ for any $x \geq 0$. What I found so far is Bernoulli's inequality, which states that $$1+x\leq\left(1+\frac{x}{n}\right)^n\xrightarrow [n\to\infty]{} e^x$$

Is there a way to utilize this proven fact in order to prove my problem?

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7 Answers 7

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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).

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  • $\begingroup$ How does this answer differ from my own? $\endgroup$
    – S.C.B.
    Commented Mar 26, 2016 at 12:49
  • $\begingroup$ @MXYMXY. I am sorry ! When I saw your answer, it seemed to me that you were treating the case of $e^x \ge x+1$ only. Did you edit your answer meanwhile ? What I did is the same as your, for sure. I just upvoted your answer. $\endgroup$ Commented Mar 26, 2016 at 12:52
  • $\begingroup$ I was just surprised that this recieved two upvotes so quickly, yet I got downvoted. Not that your answer is bad, of course. $\endgroup$
    – S.C.B.
    Commented Mar 26, 2016 at 12:55
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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 $x \geq 0.$

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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 $t$.

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  • $\begingroup$ Why the downvote? $\endgroup$
    – S.C.B.
    Commented Mar 26, 2016 at 12:52
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    $\begingroup$ Welcome to the club of victims of anonymous downvoters ! $\endgroup$ Commented Mar 26, 2016 at 12:53
  • $\begingroup$ @ClaudeLeibovici Doesn't matter, I guess. $\endgroup$
    – S.C.B.
    Commented Mar 26, 2016 at 12:54
  • $\begingroup$ If you are concerned by mathematics (and I know you are since I already had some opportunities to appreciate your answers), it does not matter at all. Continue on this way and I am sure that, one of these days, I shall see you being a very good mathematician. You deserve ! Cheers. $\endgroup$ Commented Mar 26, 2016 at 13:00
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It suffices to prove it for $0\leq x<1$ because for $x\geq 1$ we have $e^{-x}>0\geq 1-x. $

We can use $\lim_{n\to \infty}(1+x/n)^n=e^x$ for the case $ 0\leq x<1$: We have $1-x>0$ so $$e^{-x}\geq 1-x\iff e^x\leq (1-x)^{-1}.$$ $$\text {Now }\quad (1+x/n)^n=\sum_{j=0}^n (x/n)^j \binom {n}{j}=\sum_{j=0}^n x^n\cdot n^{-j}\binom {n}{j}$$ $$\text {and }\quad n^{-j}\binom {n}{j}\leq n^{-j}\;\frac {n^j}{j!}=\frac {1}{j!} \leq 1$$ $$\text {so }\quad (1+x/n)^n\leq \sum_{j=0}^nx^j\leq \sum_{j=0}^{\infty}x^j=(1-x)^{-1}.$$ Now let $n\to \infty$ and we have $e^x\leq (1-x)^{-1}.$

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Actually, the Bernoulli inequality completely proves the statement $$e^x \geq 1 +x,\ x\in\mathbb R$$

Let me write down the interesting part from the Wiki:

Let $r\geq 0$ be integer. Then $$(1+x)^r \geq 1 + rx$$ for all $x\geq-1$. Furthermore, if $r$ is even, then the above inequality holds for all real $x$.

Now, we simply conclude that $$(1+\frac x{2n})^{2n} \geq 1 + x$$ for all real $x$, and let $n\to\infty$.

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  • $\begingroup$ @Flo Ryan, in case you've seen my previous answer, but not the edit, please take a look, as this should be much more elegant. $\endgroup$
    – Ennar
    Commented Mar 26, 2016 at 13:45
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$e^{u} = \sum_{n=0}^{\infty} \frac{u^n}{n!} \geq 1 + u,$ for all $u \in \mathbb{R}.$

Choose $u = -x$ and you have the inequality.

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Remember that $e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...$

We can rewrite $1-x = e^{-x}-\frac{x^2}{2!}+\frac{x^3}{3!}-...$

So, the inequality becomes $e^{-x} \geq e^{-x}-\frac{x^2}{2!}+\frac{x^3}{3!}-..$

Which means $0 \geq -\frac{x^2}{2!}+\frac{x^3}{3!}-...$

Which is easy to see is true

EDIT: Provided $x\in[0,1]$. If $x$ is not in this interval the original inequality is trivial to see anyways because the RHS is non-positive and $e^{-x}$ is never non-positive.

Second EDIT: If you wish to use Bernoulli's inequality, simply replace $+x$ by $-x$ in the original expression. You get $1-x \leq (1-\frac{x}{n})^n \to e^{-x}$ as $n \to \infty$.

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  • $\begingroup$ The fact that $-\frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ is positive doesn't seem immediate to me. $\endgroup$
    – joachxm
    Commented Mar 26, 2016 at 12:51
  • $\begingroup$ It isn't positive. It's less than 0. $\endgroup$
    – Sorey
    Commented Mar 26, 2016 at 12:52
  • $\begingroup$ Oh, my bad, I agree with you then. $\endgroup$
    – joachxm
    Commented Mar 26, 2016 at 12:53

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