simple inequality question involving positive integers

Question

There are two numbers $x,y$ such that $x+y=N-1$ and there are also two numbers $a,b$ such that $a+b=N$

Then it is said that there can be only two cases: $x\geq a$ or $y\geq b$

All numbers $x,y,a,b,N$ are natural numbers.

I do not understand why are these two are the only possible cases. Can someone help me visualize this?

Answer 1

Both of them cannot be true because that would mean $N=a+b\leq x+y \leq N-1$, which is impossible. If neither of them is true, then $x \le a - 1$ and $y \le b - 1$, i.e., $N-1 = x+y\leq a-1+b-1=N-2$, which is again impossible.