1. ## Differentiation Problem

Let $\displaystyle a>b>0$ and let $\displaystyle n\in\mathbb{N}$ satisfy $\displaystyle n\geq 2$ . Prove that $\displaystyle a^{1/n} - b^{1/n} < (a-b)^{1/n}$

I am given the following hint:
Show that $\displaystyle f(x) := x^{1/n} - (x-1)^{1/n}$ is decreasing for $\displaystyle x\geq 1$ , and evaluate $\displaystyle f$ at $\displaystyle 1$ and $\displaystyle a/b$

2. Originally Posted by Kipster1203
Let $\displaystyle a>b>0$ and let $\displaystyle n\in\mathbb{N}$ satisfy $\displaystyle n\geq 2$ . Prove that $\displaystyle a^{1/n} - b^{1/n} < (a-b)^{1/n}$

I am given the following hint:
Show that $\displaystyle f(x) := x^{1/n} - (x-1)^{1/n}$ is decreasing for $\displaystyle x\geq 1$ , and evaluate $\displaystyle f$ at $\displaystyle 1$ and $\displaystyle a/b$
It suffices to prove this for a fixed but arbitrary $\displaystyle a,b>0$

$\displaystyle b^{\frac{1}{n}}-a^{\frac{1}{n}}<(b-a)^{\frac{1}{n}}\Leftrightarrow \frac{\left(b^{\frac{1}{n}}-a^{\frac{1}{n}}\right)^n}{b-a}=\frac{\left(\left(\frac{b}{a}\right)^{\frac{1}{ n}}-1\right)^n}{\frac{b}{a}-1}<1$.

But, the LHS is clearly $\displaystyle \frac{f\left(\frac{b}{a}\right)-f(1)}{\frac{b}{a}-1}$ where $\displaystyle f(x)=\left(x^{\frac{1}{n}}-1\right)^n$. Thus, by the mean value we have that the LHS equals $\displaystyle f'(c)=\frac{c^{\frac{1}{n}}\left(c^{\frac{1}{n}}-1\right)^{n-1}}{c}=c^{\frac{n-1}{-n}}\left(c^{\frac{1}{n}}-1\right)^{n-1}=\left(1-c^{\frac{-1}{n}}\right)^{n-1}$ for some $\displaystyle c\in\left[1,\tfrac{b}{a}\right]$. But, clearly then $\displaystyle 0\leqslant 1-c^{\frac{-1}{n}}<1\implies 0<\left(1-c^{\frac{-1}{n}}\right)^{n-1}<1$.

The conclusion follows.

3. Originally Posted by Kipster1203
Let $\displaystyle a>b>0$ and let $\displaystyle n\in\mathbb{N}$ satisfy $\displaystyle n\geq 2$ . Prove that $\displaystyle a^{1/n} - b^{1/n} < (a-b)^{1/n}$

I am given the following hint:
Show that $\displaystyle f(x) := x^{1/n} - (x-1)^{1/n}$ is decreasing for $\displaystyle x\geq 1$ , and evaluate $\displaystyle f$ at $\displaystyle 1$ and $\displaystyle a/b$

Show that $\displaystyle f'(x)<0\,\,\forall x>1$ and then $\displaystyle 1<a\slash b\Longrightarrow f(1)<f(a\slash b)$ and you're done (caution : beware of negative powers!)

Tonio