Prob. 5, Sec. 6.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: How to show this function is strictly decreasing using derivative Here is Prob. 5, Sec. 6.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

Let $a > b > 0$ and let $n \in \mathbb{N}$ satisfy $n \geq 2$. Prove that $a^{1/n} - b^{1/n} < (a-b)^{1/n}$. [Hint: Show that $f(x) \colon= x^{1/n} - (x-1)^{1/n}$ is decreasing for $x \geq 1$, and evaluate $f$ at $1$ and $a/b$.]

My Attempt:

We find that for $x > 1$,
  $$
\begin{align}
 f^\prime(x) &= \frac{1}{n}x^{\frac{1}{n} - 1} - \frac{1}{n}(x-1)^{\frac{1}{n} - 1} \\
&= \frac{1}{n} \left( \frac{x^{1/n}}{x} - \frac{(x-1)^{1/n}}{x-1} \right) \\
&= \frac{1}{nx(x-1)} \left( x^{1/n}(x-1) - x(x-1)^{1/n} \right) \\
&= \frac{x^{1/n}(x-1)^{1/n}}{nx(x-1)} \left( (x-1) - x \right) \\
&= -\frac{1}{nx^{1-\frac{1}{n}} (x-1)^{1-\frac{1}{n}} }. 
\end{align}
$$

PS (based on the answer by @auscrypt):

We find that for $x > 1$,
  $$
\begin{align}
 f^\prime(x) &= \frac{1}{n}x^{\frac{1}{n} - 1} - \frac{1}{n}(x-1)^{\frac{1}{n} - 1} \\
&= \frac{1}{n} \left( \frac{x^{1/n}}{x} - \frac{(x-1)^{1/n}}{x-1} \right) \\
&= \frac{1}{nx(x-1)} \left( x^{1/n}(x-1) - x(x-1)^{1/n} \right) \\
&= \frac{x^{1/n}(x-1)^{1/n}}{nx(x-1)} \left( (x-1)^{1-1/n} - x^{1-1/n} \right) \\
&< 0. 
\end{align}
$$
  because $1-1/n$ is positive for every $n \in \mathbb{N}$ such that $n \geq 2$ and because $0 < x-1< x$, which implies that
  $$ 0 < (x-1)^{1-1/n} < x^{1-1/n},$$
  and hence
  $$ (x-1)^{1-1/n} - x^{1-1/n} < 0.$$
Thus $f^\prime(x) < 0$ for all $x > 1$. Therefore the function $f$ is strictly decreasing on the interval $[1, +\infty)$. That is, for any $x > 1$ we have $f(x) < f(1)$.
Now if $a > b > 0$, then $a/b > 1$, and so we have 
  $$ f(a/b) < f(1),$$
  that is,
  $$ \left(\frac{a}{b}\right)^{\frac{1}{n}} - \left(\frac{a}{b} - 1\right)^{\frac{1}{n}} < 1,$$
  which amounts to
  $$ \frac{a^{1/n} - (a-b)^{1/n} }{ b^{1/n} } < 1, $$
  which implies
  $$ a^{1/n} - (a-b)^{1/n} < b^{1/n},$$
  and hence 
  $$ a^{1/n} - b^{1/n} < (a-b)^{1/n}, $$
  as required.

Are there any issues with this proof?

There is an error; the factorisation $x^{1/n}(x-1) - x(x-1)^{1/n} = x^{1/n} (x-1)^{1/n} ((x-1)-x)$ is incorrect and should be $x^{1/n}(x-1) - x(x-1)^{1/n} = x^{1/n} (x-1)^{1/n} ((x-1)^\frac{n-1}{n}-x^\frac{n-1}{n})$.
To correctly prove this, note that since $\frac{1}{n}-1$ is negative we have $x^{\frac{1}{n}-1}$ is decreasing. This implies $(x-1)^{\frac{1}{n}-1} > x^{\frac{1}{n}-1} $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.
