Can You Prove This Inequality Involving Binomial Coefficients?

[Chris L T521]

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:

Spoiler

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

 

[Chris L T521]

This week's problem was correctly answered by MarkFL and Sudharaka. You can find MarkFL's solution below.

Spoiler

I would choose to begin with:

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

Using the binomial theorem, we may write:

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

Using the identity:

$${n \choose r}={n \choose n-r}$$

there results:

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

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

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

Hence, we may state:

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

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

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

Shown as desired.