Mean Value Theorem exercise (Analysis)

antiemptyv
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Homework Statement



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):= x^{1/n}-(x-1)^{1/n} is decreasing for x\geq 1, and evaluate f at 1 and a/b.]

Homework Equations



I assume, since this exercise is at the end of the Mean Value Theorem section, I am to use the Mean Value Theorem.

The Attempt at a Solution



I can show what the hint suggests. I guess I'm not sure how those ideas help exactly.
 
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Did you evaluate f at 1 and a/b?

After you do that, keep in mind that a>b>0. You don't need to explicitly apply the mean value theorem (but you probably implicitly applied it when you proved that f(x) was decreasing for x>=1).
 
Thanks for your quick reply. Let's see what we have here...

f(1)=1

and

f(\frac{a}{b}) = \frac{a^{1/n}-(a-b)^{1/n}}{b^{1/n}}.
 
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ohh, i think i see now.

1 < a/b

and, since f is decreasing,

f(\frac{a}{b}) = \frac{a^{1/n}-(a-b)^{1/n}}{b^{1/n}} < 1 = f(1)

and the rest is just algebra to show

a^{1/n} - b^{1/n} < (a-b)^{1/n}.

look good?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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