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November 29th, 2017, 09:03 AM   #1
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Another factorial inequality

Prove that $\displaystyle n!^2 \geq n^n$ for $\displaystyle n\in N$
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November 29th, 2017, 02:45 PM   #2
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Mathematical induction can work. For n=1 or 2, equality. For n=3, 36 > 27.

In general. $(n+1)!^2=(n+1)^2(n!)^2$
$(n+1)^{n+1}=(n+1)n^n(1+\frac{1}{n})^n<(n+1)n^ne<( n+1)^2(n!)^2$, for n > 2.
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