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how do you prove this inequality? $\sqrt[n]{x}-1 \le \frac{x-1}{n}$

It looks to me as if Bernoulli's inequality would be useful.

How about the following $\frac{x-1}{n} \geq \sqrt[n]{x}-1$ From here on out I would have to show that the left hand side is $\geq (1+x)^n$ and that the right hand side $\le 1+nx$

Problem is that I don't know how to do that. Can someone help me out here? Thanks in advance.

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HINT

We have

$$\sqrt[n]{x}-1 \le \frac{x-1}{n}\iff \sqrt[n]{x} \le \frac{x+n-1}n$$

then by AM-GM

$$\frac{x+\overbrace{1+\ldots+1}^{n-1\,terms}}n\ge...$$

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  • $\begingroup$ Ah alright I think I see what you mean. $$\frac{x+\overbrace{1+\ldots+1}^{n-1\,terms}}n\ge \frac{x}{n} \ge \sqrt[n]{x}$$ $\endgroup$ – D. John Nov 11 '18 at 12:34
  • $\begingroup$ The problem is that I only know AM-GM for $$\frac{x+y}{2} \ge \sqrt[2]{xy}$$ Is $$\frac{x+y}{n} \ge \sqrt[n]{xy}$$ Correct as well? $\endgroup$ – D. John Nov 11 '18 at 12:37
  • $\begingroup$ @D.John We just need to apply the AM-GM inequality. What is the intermediate step? Note that on the LHS we have the AM of $n$ terms, one x nad $n-1$ ones. $\endgroup$ – gimusi Nov 11 '18 at 12:37
  • $\begingroup$ @D.John The general statement is $$\frac{x_1+...+x_n}{n} \ge \sqrt[n]{x_1x_2...x_n}$$ $\endgroup$ – gimusi Nov 11 '18 at 12:38
  • $\begingroup$ Alright thank you very much that really helped me out a lot! $\endgroup$ – D. John Nov 11 '18 at 12:41
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Bernoulli's inequality states that $$(1+h)^n\ge 1+nh$$ when $n\in\Bbb N$ and $h>-1$. Try $h=(x-1)/n$.

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hint

Put $$f(x)=x^\frac 1n$$

$f$ is differentiable at $(0,+\infty)$, and by MVT

$$f(x)-f(1)=(x-1)f'(c)$$

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There is a formulation of Bernoulli's inequality which gives a very short proof: $$\forall y> 0,\quad y^n-1\ge n(y-1)$$

Now the inequality you have to prove can be written as $$x-1\ge n(\sqrt[n]{x}-1).$$ So just set $\;y=\sqrt[n]{x}$.

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