Showing an Inequality Holds True

Question

If $x$ is a positive real number and $n$ is a positive integer, prove the inequality, $\sqrt[n]{1+x}-1 \le \frac{x}{n}$.

I tried to show this was true by doing the following:

For $n = 1$, $\sqrt[1]{1+x}-1 = x \le \frac{x}{1} = x$

Assuming the proposition holds true for all positive integers in the interval $[1,n]$

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

My issue is that I don't know how to show that $\sqrt[n+1]{1+x}-1 \le\frac{x}{n+1}$. I definitely feel like there is a better approach, but to be honest I haven't faced many problems like this. This problem was given in Donald Knuth's Art of Computer Programming: Volume 1 Third Edition.

Thanks for any help.

Answer 1

I think there's a much more elegant solution.

Using the binomial theorem, we have:

$$ \begin{align} \Big(1 + \frac{x}{n} \Big)^n &= \sum_{i = 0}^n {n \choose i} \Big(\frac{x}{n}\Big)^i \\ &= 1 + x + \sum_{i = 2}^n {n \choose i} \Big(\frac{x}{n}\Big)^i \\ &\geq 1 + x \end{align} $$ where the last inequality stems from the fact that all terms in the sum are positive.

Thus, we easily see that

$$\Big(1 + \frac{x}{n} \Big)^n \geq 1 + x$$

And rearranging, we obtain:

$$\frac{x}{n} \geq \sqrt[n]{(1 + x)} - 1$$

Which is what we wanted to show.

Answer 2

It's just the Bernoulli's inequality: $$\sqrt[n]{1+x}=(1+x)^{\frac{1}{n}}\leq1+\frac{1}{n}\cdot x=1+\frac{x}{n}.$$