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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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    Yes. But I wanted to see what happends with$x$arround zero.2011-11-30

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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.