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

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.

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

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

• Ah alright I think I see what you mean. $$\frac{x+\overbrace{1+\ldots+1}^{n-1\,terms}}n\ge \frac{x}{n} \ge \sqrt[n]{x}$$ – D. John Nov 11 '18 at 12:34
• 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? – D. John Nov 11 '18 at 12:37
• @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. – gimusi Nov 11 '18 at 12:37
• @D.John The general statement is $$\frac{x_1+...+x_n}{n} \ge \sqrt[n]{x_1x_2...x_n}$$ – gimusi Nov 11 '18 at 12:38
• Alright thank you very much that really helped me out a lot! – D. John Nov 11 '18 at 12:41

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

hint

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

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

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

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