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Bernoulli's inequality says that $(1+x)^n \geq 1+nx$ for all $x > -1$ and for all $n \in \mathbb{N}$.

The questions asks $what \ can \ you \ say \ if \ x\le-1 \ ?$

So I was just trying out different numbers of $x$ that is less that $-1$ and for all of the $x$ and $n$ that I have tried, the inequality seems to apply for $x > -2$ for all $n \in \mathbb{N}$, and the inequality does not hold for $x<-2$.

I am just wondering if this is the case. If it is, is there a specific proof that I can go through in order to show it.

Thanks to anybody who helps.

Lucy
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1 Answers1

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I used some brute force, but I think it can be considered as a solution.

If $n=2k+1, x = -1-t, ~0\le t \le 1$, then $(-1)^nt^n \ge 1-n(t+1)$ as it can be rewritten in the way:

$$-t^{2k+1} \ge -2k(t+1)-t ~~\rightarrow~~ t^{2k+1} \le 2k(t+1)+t$$ The last inequality can be proved by considering functions $f = t^{2k+1}$ and $g = 2k(t+1)+t$ with their derivatives on [0,1]:

$$f' = (2k+1)t^{2k} \le (2k+1) = g' ~\text{and} ~~f(0)\le g(0).$$

If $n = 2k, x = -1-t, ~0\le t \le 1$ we have $$t^{2k} \ge 1-2k(t+1).$$

This inequality is even easier as left part is always positive and the right is not. Hope that I haven't made a mistake in this.

Andrei Kulunchakov
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