if $x\le y+z$ for every $z>0$ then $x\le y$

Question

Here's what I have to show:

If $x\le y+z$ for every $z>0$ then $x\le y$.

I tried proof by contradiction but it does not seem to work. Suppose $y<x$. Then $x-y>0$. Let $z_{0}=x-y$. Then $x\le y+z_{0} \Rightarrow x\le x$ which is true. Had $x$ been strictly less than $y+z$, it would have worked.

Can I get hints?

Answer 1

"Had x been strictly less than y+z, it would have worked."

So instead of making $z = x-y$ make $0 < z < x-y$.

Then $x \le y + z < y + x-y = x$.

Answer 2

You are almost there. Assume that $x\gt y$, then set $z=\frac{x-y}2\gt0$. That would mean that $x=y+2z\gt y+z$ and $z\gt0$, which is a contradiction.