# Math Help - Differentiation Problem

1. ## Differentiation Problem

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

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

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

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

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

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

The conclusion follows.

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

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

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

Tonio
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