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

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#### Chris L T521

##### Well-known member
Staff member
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 $y$.

#### Chris L T521

##### Well-known member
Staff member
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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