I need help with the following problem.
For $k,x, n > 0$ such that $k + x < n$, prove that $$\left ( \frac{n-k-x}{n-x} \right )^x \leq \binom{n-x}{k} \binom{n}{k}^{-1}.$$
I've tried using various bounds such as $e^{-t} > 1-t > e^{-t - t^2/2}$, the standard bounds for binomial coefficients, etc.
Any solutions/suggestions would be appreciated.
The RHS of your inequality can be written as $$\begin{align}\frac{(n-x)!}{k!(n-x-k)!}\cdot\frac{k!(n-k)!}{n!} &=\frac{(n-k)!}{(n-x-k)!}\div \frac{n!}{(n-x)!} \\\\&=\frac{(n-k)(n-k-1)\cdots (n-x-k+1)}{n(n-1)\cdots (n-x+1)} \\\\&=\frac{n-k}{n}\times \frac{n-k-1}{n-1}\times\cdots\times \frac{n-x-k+1}{n-x+1} \\\\&=\prod_{j=1}^{x}\frac{n-k-j+1}{n-j+1}\end{align}$$
So, your inequality is equivalent to $$\left ( \frac{n-k-x}{n-x} \right )^x \leq \prod_{j=1}^{x}\frac{n-k-j+1}{n-j+1}\tag1$$
In order to prove that $(1)$ holds, it is sufficient to prove that $$\frac{n-k-x}{n-x}\leq \frac{n-k-j+1}{n-j+1}\tag2$$ holds for every $j$ such that $1\le j\le x$.
We see that $$\begin{align}(2)&\iff (n-k-x)(n-j+1)\leq (n-x)(n-k-j+1) \\\\&\iff j\le x+1\end{align}$$ which holds for every $j$ such that $1\le j\le x$.
Hence, we see that your inequality holds.