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For which $x$ the following inequality is true: $$(1-x)^n\leq 1-\frac{nx}{2},$$ where $n$ is natural number?

This is inequality opposite to Bernoulli inequality. For sure $x$ should be between zero and $1$. $x=0$ is fine. But what is the upper bound for $x$?

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This is what I can get so fare: (1-x)^n<=e^{-nx}<=1-\frac{nx}{2}. Let y=e^{-nx}+\frac{nx}{2}-1<=0. y'=-ne^{-nx}+\frac n2=0 if x=\frac{ln 2}{n}. So, x \in [0, \frac{ln 2}{n}]. –  David Nov 30 '11 at 5:11
    
Why should $x$ be less than 1? The lhs will go to $-\infty$ as $x$ goes to $\infty$ much faster than the rhs. Thus, for a high enough value of $x$ the inequality will hold. –  tards Nov 30 '11 at 5:13
    
Yes. But I wanted to see what happends with x arround zero. –  David Nov 30 '11 at 5:16

1 Answer 1

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For $n=1$ it is true for all $x \ge 0$. If you are interested in $x$ close to $0$, you might think to ignore the terms in $x^3$ and higher, getting $1-nx+\frac{n(n-1)}{2}x^2 \le 1-\frac{nx}{2}$ or $x \lt \frac{2}{n-1}$, but that is not so close. For $n=4$ the solution is about $x \lt 0.456311$, for $n=6$, $x \lt 0.291$. Alpha makes quick work of these.

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