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I want to show the below statement.

$n!x^{n!} \leq n\cdot x^{n}$

Where $0 \leq x <1$ for all $n\in \mathbb{N}$

My approach is induction.

The Induction start

This is easy. Set n=1 and the statement $n!x^{n!} \leq n\cdot x^{n}$ is true.

Induction step

Now I assume that $(n-1)!x^{(n-1)!} \leq (n-1)\cdot x^{n-1}$ is true. But I'm stuck here. I have tried to deduce the statement $n!x^{n!} \leq n\cdot x^{n}$ but my attempts seem to fail..

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Another way to see this cannot be true: Fix $n$. Suppose the inequality is true for $0\le x<1$. Take limit $x\to 1^-$. Then (using continuity) you get $n!\le n$. – Martin Sleziak May 27 '12 at 3:52
up vote 5 down vote accepted

The inequality is not true. For instance, for $n=3$, we get that $$3! x^{3!} \leq 3 x^3$$ $$6x^6 \leq 3 x^3$$ $$x^3 \leq \frac12$$ The above is true only for $$x \in \left[0,\frac1{2^{1/3}} \right]$$

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Thank you! It explains why it was so hard to prove. – characters May 26 '12 at 21:58

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