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I need to prove that $\frac{1}{1-x}\geq e^x$ for all $x<1$. I've allready proved that $1+x\leq e^x$ for all $x\in\mathbb{R}$. I've had some other ideas but I'm not allowed to use anything to different from basic analysis of functions and series, i.e. no differentials or integrals.

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Let $y=-x$

$$\frac{1}{1-x}\geq e^x \iff \frac{1}{1+y}\geq e^{-y}\iff e^y \geq 1+y$$

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    $\begingroup$ Thanks. I think I was to deep inside to see this. $\endgroup$ – blub Dec 14 '17 at 21:21
  • $\begingroup$ When you have negative variable the first thing to do is alway convert in positive! $\endgroup$ – gimusi Dec 14 '17 at 21:22
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You have $1+x\leq e^x$ so $\dfrac{1}{1+x}\geq e^{-x}$, let $y=-x<0$ then $$e^y\leq\dfrac{1}{1-y}$$

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    $\begingroup$ @gimusi Thanks. Actually I didn't know what to prove! $\endgroup$ – Nosrati Dec 14 '17 at 21:34
  • $\begingroup$ You are welcome! Bye. $\endgroup$ – gimusi Dec 14 '17 at 21:35

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