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I am trying to prove: For $n>1$

$$n^{n-1}\geq 1^{n-1}+2^{n-1}+\cdots+\left( n-1\right) ^{n-1}$$

I am quite sure that this is correct. (checked with several arbitrary n values) But, no idea how to prove it. Any comments appreciated.

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up vote 3 down vote accepted

Consider that, for any $n\geq 2$, $f(x)=x^{n-1}$ is an increasing function on $\mathbb{R}^+$, so:

$$1^{n-1}+2^{n-1}+\ldots+(n-1)^{n-1}\leq\sum_{k=1}^{n-1}\int_{k}^{k+1}f(x)\,dx\leq\int_{0}^{n}x^{n-1}dx = n^{n-1}.$$

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It might be useful to add the details on why the initial inequality works (i.e., that each term $i^{n-1}$ underestimates the piece of the integral from $i$ to $i+1$, in standard Riemann box-sum fashion). – Steven Stadnicki Nov 26 '12 at 16:48
Thanks a ton. Beautiful and easy proof. Can you also check the other question I posted? Proof should be similar to this one but does not trivially extend, I guess. – Emre Nov 26 '12 at 21:37
@Emre: I gave a proof to your other inequality, where I used midpoint-convexity to strengthen this result. – Jack D'Aurizio Nov 27 '12 at 9:30

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