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Suppose $m>x$, where $x$ belongs to positive Real Numbers and letting $n\ge m$, show that $${x^n\over n!}\le{x^n\over m^n}\cdot{m^m\over (m-1)!}$$ Frankly, I consider myself as a beginner in Inequalities and therefore, I don't know how to approach such problems. So,in essence I have two main questions for the community:

1) How to solve the above written problem?

2) How to check whether one's proof is correct or not? (This is major issue as I have often made gross errors while proving inequalities, even though I felt quite convinced with my argument)

Any inputs and suggestions will be much appreciated.

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  • $\begingroup$ Is it $x^m$ on the right side? $\endgroup$ – Jimmy R. Jan 31 '16 at 12:23
  • $\begingroup$ My book mentions $x^n$ on the right side.. $\endgroup$ – model_checker Jan 31 '16 at 12:24
  • $\begingroup$ So, $x^n$ just cancels out from both sides? $\endgroup$ – Jimmy R. Jan 31 '16 at 12:26
  • $\begingroup$ Yeah, I guess... $\endgroup$ – model_checker Jan 31 '16 at 12:27
  • $\begingroup$ @Shrey Aryan you should accept any one answer if your query has been answered ...if not you can ask for more information or clear your confusions by adding a comment below any answer... $\endgroup$ – Freelancer Jan 31 '16 at 15:12
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I suppose $m,n$ are natural numbers. This inequality is equivalent to $\;n!\ge m^{n-m} (m-1)!$

Now, as $n\ge m$, we can rewrite $n!$ as $$n!=(1\cdot2\cdots(m-1))(\underbrace{m(m+1)\cdots n}_{\text{each factor is }\ge~ m})\ge\begin{cases}(m-1)!\; m^{n-m+1} \ge (m-1)!\; m^{n-m}\\[1ex] m!\;m^{n-m}\end{cases} $$

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In such kind of problems start with cancelling the common terms on both the sides since in this case it is $x^n$ which is positive so you can demolish that term without any mercy ...also notice that the RHS can also be reduced so we can write this as (new inequality) $$\frac{1}{n!}\le\left(\frac{1}{m^{n-m-1}×m×\left((m-1)!\right)}\right)$$ or this can also be written as $$\frac{1}{n!}\le\left(\frac{1}{m^{n-m-1}×(m!)}\right)$$ now I think you can easily solve the problem as you can easily compare to prove the inequality also we know that $(n!)\le(m!)$

Edit:as pointed out in other answers this is obviously same as the inequality $$\;n!\ge m^{n-m-1}m!$$ which has also been solved ...good for you...and about the part how you can check whether it is true or not...well most of the begginers like me just use basic principle of mathematical induction which is very easy to use...but then again I use it only for natural numbers...but the question here is a more general case ....that is for real numbers ...but still since $N$ is a subset of $R^+$ so it is a fine proof checker at least for a beginner...also another thing you should take care is while multiplying in an inequality as I also used to neglect reversing the inequality whenever i used to multiply negative numbers on both the sides of the inequality ....also the inequalities on logarithm will be a nightmare in the beginning but you will get used to them overtime...also I would like to advice you to try solving more and more questions on your own instead of just asking a question here which doesn't show any effort from your side in solving the question...best of luck..

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As written, is equivalent to the obvious inequality $$(m-1)!\, m^{n-m} \le (m-1)!m(m+1)\cdots n = n!.$$

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I can't leave a comment yet, so here's a maybe incomplete answer: Assuming that $n$ and $m$ are natural numbers, the inequation can be rewritten as:

$m^{n-m-1} \le (m+1)*(m+2)*...*n$

which is simply stricter than

$m^{n-m} \le (m+1)*(m+2)*...*n$

$n-m$ is non negative and the left hand side has the same amount of factors ($n-m$ times $m$) as the right hand side, which has $n-m$ times a number $> m$.

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