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I've seen the following and I'm not sure whether it is true or not, and if yes, why it holds.

$(1-p)^x \geq 1-p x$ for $p\in (0,1)$ and $x>0$.

Do I need some additional Information to prove that?

We have $$(1-p)^x = \sum_{n=0}^{\infty} { x \choose n} \cdot (-p)^n=1-px + \sum_{n=2}^{\infty} { x \choose n} \cdot (-p)^n$$.

Now we would need to Show that $$\sum_{n=2}^{\infty} { x \choose n} \cdot (-p)^n \geq 0$$, then everything would be great. But I'm not sure whether this is possible.

Thank you very much for your help

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  • $\begingroup$ Have you heard of Newton's generalization of the binomial theorem? $\endgroup$ – Guy Apr 3 '14 at 12:09
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    $\begingroup$ The first inequality above, for $\;x\in\Bbb N\;$ is called "Bernoulli's Inequality" and can be easily proved by induction. $\endgroup$ – DonAntonio Apr 3 '14 at 12:12
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    $\begingroup$ You may have intended Bernoulli's Inequality, but the statement you gave is incorrect. Check for e.g. with $p=\frac12, x = \frac12$. The correct statement is here: en.wikipedia.org/wiki/Bernoulli's_inequality#Generalization $\endgroup$ – Macavity Apr 3 '14 at 12:18
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It's false for $x=p=\frac{1}{2}$. However, it's true for $x\geq1$. Indeed, you can study $p\rightarrow(1-p)^x-1+px$, the derivative is positive.

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  • $\begingroup$ actually I thought of $x$ being integer, so implicitly assuming $x\geq 1$ when I wrote $x>0$. sorry for the confusing/wrong writing.. $\endgroup$ – user136457 Apr 3 '14 at 13:51

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