Mean Value Theorem exercise (Analysis)

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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].
 
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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?
 
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