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I wish to prove the following inequality. I have done a plot and it looks ok. $\mu \geq 1$ is an integer and so is $h$, where $0 \leq h \leq \mu-1$.

I started like this:

$\frac{e^{-\mu}\mu^{(\mu+h)}}{(\mu+h)!} \geq \frac{e^{-\mu}\mu^{(\mu-h-1)}}{(\mu-h-1)!}$.

I have rewritten this to

$(\mu-h-1)!\mu^{(\mu+h)} \geq (\mu+h)!\mu^{(\mu-h-1)}$,

however I am not able to go further from there. Can you help me out?

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    What inequality are you trying to prove?2010-10-11

1 Answers 1

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Cancel a factor of $\mu^\mu$ on both sides, move the powers of $\mu$ to the left, and move the factorials to the right to obtain the equivalent statement

$$\mu^{2h+1}\geq (\mu+h)(\mu+h-1)(\mu+h-2)\cdots(\mu-h+1)(\mu-h).$$

Break up the right hand side into the middle term, $\mu$, along with $h$ pairs of the form $(\mu+k)(\mu-k)$. Since $(\mu+k)(\mu-k)=\mu^2-k^2\lt\mu^2$ for $k>0$, it should now be clear.

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    I was going to invoke the AM-GM inequality in my solution, but your elementary argument is really nice.2010-10-11
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    @Rahul: Thank you. Perhaps your argument would help the OP learn a more generally applicable technique.2010-10-11
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    Thank you, very nice answer. I would not have thought of this myself.2010-10-11
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    You're welcome. It becomes much clearer with additional simplification. I wouldn't have thought of it either if it weren't second nature to combine powers and see if dividing factorials would simplify things. Once I saw the product of 2h+1 terms on each side, and the particular pattern on the right, the inequality a^2 > (a+b)(a-b) jumped out at me.2010-10-11