If $a>b>0$ and $n\in\mathbb{N}$ then $a^{1/n}-b^{1/n}<(a-b)^{1/n}$
I know $f(x)=x^{1/n}-(x-1)^{1/n}$ is strictly decreasing if $x>1$. Since after taking the derivative I get that $f'(x)=\frac{1}{n}x^{\frac{1-n}{n}}-\frac{1}{n}(x-1)^{\frac{1-n}{n}}$ which is decreasing for all $x>1$
Now evaluating $f(1)$ I get just $1$ and evaluating $f\left(\frac{a}{b}\right)$ I get $\frac{a}{b}^{\frac{1}{n}}-\left(\frac{a-b}{b}\right)^{\frac{1}{n}}$. Now I'm not sure how to then show that $a^{1/n}-b^{1/n}<(a-b)^{1/n}$