Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question
1  
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
add comment

1 Answer

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 $$x \in \left[0,\frac1{2^{1/3}} \right]$$

share|improve this answer
3  
Thank you! It explains why it was so hard to prove. –  characters May 26 '12 at 21:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.