X Choose n leq y Choose n + z Choose n-1 where n leq y leq x leq y+1 and n-1 leq z leq y

Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.

Let $x, y \in \R$ be real numbers which satisfy:
 * $n \le y \le x \le y + 1$

Let $z$ be the unique real number $z$ such that:


 * $\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$

where $n - 1 \le z \le y$.

Its uniqueness is proved at Uniqueness of Real $z$ such that $\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$.

Then:
 * $\dbinom x n \le \dbinom y n + \dbinom z {n - 1}$

Proof
If $z \ge n$, then from Ordering of Binomial Coefficients:


 * $\dbinom z {n + 1} \le \dbinom y {n + 1}$

Otherwise $n - 1 \le z \le n$, and:


 * $\dbinom z {n + 1} \le 0 \le \dbinom y {n + 1}$

In either case:


 * $(1): \quad \dbinom z {n + 1} \le \dbinom y {n + 1}$

Therefore:

and so $x \ge z + 1$.

Now we are to show that every term of the summation:
 * $\displaystyle \binom x {n + 1} - \binom y {n + 1} = \sum_{k \mathop \ge 0} \dbinom {z - k} {n - k} t_k$

where:
 * $t_k = \dbinom {x - z - 1 + k} {k + 1} - \dbinom {y - z - 1 + k} {k + 1}$

is negative.

Because $z \ge n - 1$, the binomial coefficient $\dbinom {z - k} {n - k}$ is non-negative.

Because $x \ge z + 1$, the binomial coefficient $\dbinom {x - z - 1 + k} {k + 1}$ is also non-negative.

Therefore:
 * $z \le y \le x$

implies that:
 * $\dbinom {y - z - 1 + k} {k + 1} \le \dbinom {x - z - 1 + k} {k + 1}$

When $x = y$ and $z = n - 1$ the result becomes:


 * $\dbinom x n \le \dbinom x n + \dbinom {n - 1} {n - 1}$

which reduces to:
 * $\dbinom x n \le \dbinom x n + 1$

which is true.

Otherwise:

This is less than or equal to:

Hence the result.