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Problem of the Week #67 - July 8th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Prove that
\[\left(\frac{m}{m+n}\right)^m \left(\frac{n}{m+n}\right)^n {{m+n}\choose m}<1\]
for all $m,n\in\mathbb{Z}^+$.

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Hint:
Consider the term for $k=m$ in the binomial theorem expansion for $(x+y)^{m+n}$ for appropriate $x$ and $y$.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by MarkFL and Sudharaka. You can find MarkFL's solution below.

I would choose to begin with:

\(\displaystyle 1=\left(\frac{m+n}{m+n} \right)^{m+n}=\left(\frac{m}{m+n}+\frac{n}{m+n} \right)^{m+n}\)

Using the binomial theorem, we may write:

\(\displaystyle 1=\sum_{k=0}^{m+n}{m+n \choose k}\left(\frac{m}{m+n} \right)^{m+n-k}\left(\frac{n}{m+n} \right)^k\)

Using the identity:

\(\displaystyle {n \choose r}={n \choose n-r}\)

there results:

\(\displaystyle 1=\sum_{k=0}^{m+n}{m+n \choose m+n-k}\left(\frac{m}{m+n} \right)^{m+n-k}\left(\frac{n}{m+n} \right)^k\)

\(\displaystyle 1=\sum_{k=0}^{n-1}{m+n \choose m+n-k}\left(\frac{m}{m+n} \right)^{m+n-k}\left(\frac{n}{m+n} \right)^k+\)

\(\displaystyle {m+n \choose m}\left(\frac{m}{m+n} \right)^{m}\left(\frac{n}{m+n} \right)^n+\sum_{k=n+1}^{m+n}{m+n \choose m+n-k}\left(\frac{m}{m+n} \right)^{m+n-k}\left(\frac{n}{m+n} \right)^k\)

Hence, we may state:

\(\displaystyle {m+n \choose m}\left(\frac{m}{m+n} \right)^{m}\left(\frac{n}{m+n} \right)^n=\)

\(\displaystyle 1-\left(\sum_{k=0}^{n-1}{m+n \choose m+n-k}\left(\frac{m}{m+n} \right)^{m+n-k}\left(\frac{n}{m+n} \right)^k+\sum_{k=n+1}^{m+n}{m+n \choose m+n-k}\left(\frac{m}{m+n} \right)^{m+n-k}\left(\frac{n}{m+n} \right)^k \right)<1\)
Thus, we may conclude:

\(\displaystyle {m+n \choose m}\left(\frac{m}{m+n} \right)^{m}\left(\frac{n}{m+n} \right)^n<1\)

Shown as desired.
 
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