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