A simple binomial inequality seeking its proof from the book? Let $k\in \mathbb{N}$ and $1\leq a \leq k$ be fixed. Imagine we have two groups $G_{1}, G_{2}$ of $k$ indistinguishable elements each forming two sequences with indices $1, \dots , k$. Choose the element with index $a$ in $G_{1}$ and the element with index $b$ in $G_{2}$. Consider the expression
$$f(k,a,b) = \binom{a+b-2}{a-1}\binom{2k-b-a}{k-a},$$
which counts the number of ways to merge the elements before the two chosen elements and after them into two sequences.
Example: $k=5$, $a=3$, $b=4$:
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The number of ways to merge the set of elements before the two chosen is $\binom{2+3}{2}$ and to merge the set of elements after the two chosen is $\binom{2+1}{1}$. In total, we get $\binom{2+3}{2}.\binom{2+1}{1}$.
Show that
$$\binom{a+b-2}{a-1}\binom{2k-b-a}{k-a}\geq min\left(\binom{2k-1-a}{k-a},\binom{a+k-2}{a-1}\right),$$
i.e.,
$$f(k,a,b)\geq min\left(f(k,a,1),f(k,a,k)\right).$$
In other words, when $k$ and $a$ are fixed, $f$ reaches its minimum at one of the endpoints $b=1$ or $b=k$.

An ugly proof by expansion of the binomial coefficients exist, but
does anyone know an elegant proof?
 A: By expanding the binomials, it can be shown directly that
$$
\frac{f(k,a,b)}{f(k,a,b-1)} = \frac{\binom{a+b-2}{a-1}\binom{2k-b-a}{k-a}}{\binom{a+b-3}{a-1}\binom{2k-b-a+1}{k-a}} \\
= \frac{a+b-2}{b-1} \frac{k-b+1}{2k-a-b+1} = (1 + \frac{a-1}{b-1})(1 - \frac{k-a}{2k-a-b+1})
$$
Interpreting this ratio as a function of a continuous $b$, it is straightforward to calculate that the ratio becomes $1$ for $b^* = \frac{ak - 1}{k-1}$.
Since $ b < 2k - a +1$, both terms in the last product always fall with $b$. It follows that the ratio is smaller than $1$ for  $ \frac{ak - 1}{k-1} < b$ and hence $f(k,a,b)$ is falling in that range of $b$, so  the minimum is obtained at the "endpoint", for the highest $b$. Conversely, for $ \frac{ak - 1}{k-1} > b$ the ratio is larger than $1$ and hence $f(k,a,b)$ is rising in that range of $b$, so  the minimum is obtained at the other "endpoint", for the lowest $b$.
Together, this proves the claim that $f(k,a,b)\geq \min\left(f(k,a,1),f(k,a,k)\right).$ $\qquad \Box$
