How to prove that $a\le b$ , if $ a0$? How to prove that $a\le b$ , if $ a<b+c$ for each $c>0$?
I tried to prove it with the reductio ad absurdum method and with the trichotomy property of two real numbers $a,b$ : $a=b$ ,$a<b$ or $a>b$. But I didn't make it.
Any advice would be helpful !

Suppose $a>b$, $(a-b)>0, a<b+(a-b)=a$ contradiction.
