prove a general inequality between radicals

Question

I've been trying to work on a proof for this for a few days now, but can't seem to see the answer. It might be trivial but I would be grateful if anyone could let me know their strategy (if not a proof) for this:

$$\sqrt[n]{n!} \lt \sqrt[n+1]{(n+1)!}$$

Thanks in advance!

Take the logarithm, then this is equivalent to $$\dfrac1n\sum_{k=1}^n\log k\le\dfrac1{n+1}\sum_{k=1}^{n+1}\log k.$$ In other wards average of $\log $ of first $n$ positive integers is an increasing function of $n.$ See this for a graphical interpretation — Bumblebee, Oct 30 '17 at 01:08
Another one: https://math.stackexchange.com/questions/1892518/prove-nth-root-of-n-is-less-than-n1-th-root-of-n1-sqrtnn-lt-sq. — Martin R, Oct 30 '17 at 06:22

score 1 · Accepted Answer · answered Oct 30 '17 at 00:03

1

Raise both side to the power of $n$ and then riase it to the power of $n+1$ and we have \begin{eqnarray*} (n!)^{n+1} < ((n+1)!)^n. \end{eqnarray*} Now cancel $(n!)^n$ \begin{eqnarray*} n! < (n+1)^n. \end{eqnarray*} This is obvious, multiply the following inequalities $1 <n+1,2<n+1, \cdots , n<n+1$.

answered Oct 30 '17 at 00:03

Donald Splutterwit

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Ah, of course, simplifying it makes it so obvious! Thank you so much! :) – mathnoob Oct 30 '17 at 00:44

prove a general inequality between radicals

1 Answers1