Mean Value Theorem exercise (Analysis)

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Homework Help Overview

The problem involves proving an inequality related to the Mean Value Theorem, specifically concerning the expression a^{1/n} - b^{1/n} and its relationship to (a-b)^{1/n} for positive numbers a and b, with n being a natural number greater than or equal to 2.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss evaluating the function f(x) = x^{1/n} - (x-1)^{1/n} at specific points and consider its decreasing nature for x ≥ 1. There is uncertainty about how these evaluations contribute to proving the original inequality.

Discussion Status

Participants are actively engaging with the problem, evaluating the function at key points and discussing implications of its decreasing behavior. Some guidance has been offered regarding the relevance of the Mean Value Theorem, though no consensus has been reached on the final steps of the proof.

Contextual Notes

There is an emphasis on the conditions a > b > 0 and the requirement that n is a natural number greater than or equal to 2. The discussion reflects an exploration of the implications of these constraints on the problem.

antiemptyv
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Homework Statement



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

[tex]f(1)=1[/tex]

and

[tex]f(\frac{a}{b}) = \frac{a^{1/n}-(a-b)^{1/n}}{b^{1/n}}[/tex].
 
Last edited:
ohh, i think i see now.

[tex]1 < a/b[/tex]

and, since f is decreasing,

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

and the rest is just algebra to show

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

look good?
 
Last edited:

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