This question was initially posted on math.stackexchange.com but did not receive any answers for half a week.

While analyzing the properties of an algorithm I am working on (I'm a computer scientist), I came up with the following inequality. I am counting the occurrences of two different events using inclusion-exclusion terms. Then, I want to prove that one occurs at least as frequently as the other.

For $0 \leq i \leq j < k$ (natural numbers), the inequality boils down to $$ \sum_{a = 1}^{i} (-1)^{a+1} \frac{a}{k - a} \binom{k}{a} \binom{k - a}{k - i} \binom{k - a}{k - j} \geq 0 \ \text{.} $$ This is also equivalent to $$ \sum_{a=0}^{i-1} (-1)^a \frac{(k-a-2)!}{a! (i-a-1)! (j-a-1)!} \geq 0 \ \text{,} $$ using common identities.

Example: If we look at $k=10$, $j=8$, and vary $i \in \lbrace 0, ..., 8 \rbrace$, we get the following summands for the first expression:

```
k=10, i=0, j=8: 0 = []
k=10, i=1, j=8: 40 = [40]
k=10, i=2, j=8: 45 = [360, -315]
k=10, i=3, j=8: 0 = [1440, -2520, 1080]
k=10, i=4, j=8: 0 = [3360, -8820, 7560, -2100]
k=10, i=5, j=8: 0 = [5040, -17640, 22680, -12600, 2520]
k=10, i=6, j=8: 0 = [5040, -22050, 37800, -31500, 12600, -1890]
k=10, i=7, j=8: 0 = [3360, -17640, 37800, -42000, 25200, -7560, 840]
k=10, i=8, j=8: 0 = [1440, -8820, 22680, -31500, 25200, -11340, 2520, -180]
```

**Edit:** The original problem that I want to tackle is
$$
\sum_{a = 1}^{i} (-1)^{a+1} \binom{k}{a} \binom{k - a}{k - i} \binom{k - a}{k - j} \leq \binom{k}{i} \binom{k}{j} \ \text{.}
$$
Using Bonferroni's inequalities, we have
$$
\sum_{a = 1}^{i} (-1)^{a+1} \binom{k}{a} \binom{k - a}{k - i} \binom{k - a}{k - j} \leq \sum_{a = 1}^{1} (-1)^{a+1} \binom{k}{a} \binom{k - a}{k - i} \binom{k - a}{k - j} = k \binom{k-1}{i-1} \binom{k-1}{j-1} = \frac{ij}{k} \binom{k}{i} \binom{k}{j} \nleq \binom{k}{i} \binom{k}{j} \ \text{.}
$$
Unfortunately, $\frac{ij}{k} \nleq 1$.

I would appreciate any suggestions or hints. Thanks in advance.