Prove that $\sqrt[n]{x}-1 \le \frac{x-1}{n}$

Question

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.

Answer 1

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

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

Answer 2

hint

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

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

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

Answer 3

Bernoulli's inequality states that $$(1+h)^n\ge 1+nh$$ when $n\in\Bbb N$ and $h>-1$. Try $h=(x-1)/n$.

Answer 4

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 $\;y=\sqrt[n]{x}$.