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Prove $\sqrt[n]{n!}\le{{n+1}\over2}$, $n \in N^+$, using induction.

This is how far I got, but then I got stuck:

$$\sqrt[n+1]{(n+1)!}\le{{n+2}\over2}$$ $$((n+1)!)^{1\over{n+1}}\le{n\over2}+1$$ $$((n+1)\times n!)^{1\over{n+1}}\le{n\over2}+1$$ $$(n+1)^{1\over{n+1}}\times n!^{1\over{n+1}}\le{n\over2}+1$$ Can i somehow substitute $(n+1)^{1\over{n+1}}$ with ${{{n+2}\over2}}$ using the original proposition $\sqrt[n]{n!}\le{{n+1}\over2}$?

How would I continue?

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A proof without induction is to note that this follows from the inequality between the arithmetic and the geometric means, applied to $1,2,\dots,n$. – Andrés E. Caicedo Nov 5 '11 at 3:14
You have started out with the thing you are trying to prove, which is a good way to make a mistake. You have to start with the statement for $n$, and deduce the statement for $n+1$. Working in the other direction is OK only if you can show that all the steps are reversible. – Gerry Myerson Nov 5 '11 at 3:18
up vote 5 down vote accepted

Start from $$\root n\of{n!}\le{n+1\over2}$$ which is

$$n!\le(n+1)^n/2^n$$ Multiply by $n+1$:

$$(n+1)!\le(n+1)^{n+1}/2^n$$ Need to show $$(n+1)^{n+1}/2^n\le(n+2)^{n+1}/2^{n+1}$$ This is $$2\le((n+2)/(n+1))^{n+1}$$

Write the right side as $(1+r)^{n+1}$ with $r=1/(n+1)$. Can you show $$(1+r)^{n+1}\ge1+(n+1)r$$

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