How can one show $\displaystyle\frac{(m+n-1)!}{m ! (n-1)!}\leq \left[\frac{e (m+n-1)}{n}\right]^{n-1}$ ?
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Denote the quotient of the right and left hand sides, $$ f(m,n)=\biggl(\frac{e(m+n-1)}n\biggr)^{n-1}\bigg/\binom{m+n-1}m. $$ Then $f(m,1)=1$ for all $m\in\mathbb N$ and $$ \frac{f(m,n+1)}{f(m,n)} =\frac{e}{\biggl(1+\dfrac1n\biggr)^n}\cdot\biggl(1+\frac1{m+n-1}\biggr)^{n-1} > 1, $$ that is, $$ f(m,n+1)> f(m,n)>\dots> f(m,2)> f(m,1)=1. $$ This proves the required inequality. |
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Another immediate proof can be obtained from $$ \frac{(m+n-1)!}{m!}\le(m+n-1)^{n-1} $$ (which is obvious) and $$ \left(\frac{n}{e}\right)^{n-1}\le(n-1)! $$ which after multiplying by $n$ and taking logs becomes $$ n\log(n)-n+1\le\sum_{k=2}^{n}\log(k) $$ which is immediate as the RHS is an obvious upper bound for $$ \int_1^n\log(x)\,dx=(x\log(x)-x)|_1^n=n\log(n)-n+1. $$ |
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