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 !
 A: Suppose $a > b$. Let $d = a - b > 0$. Then $a = b + d$. But this contradicts the assumption that $a < b +d$ for all positive $d$. 
A: Suppose $a>b$, then $a-b>0$. Let $c=b-a$ and we have
$$a<b+(a-b)=a$$
We found a contradiction.
A: Suppose $a>b$, $(a-b)>0, a<b+(a-b)=a$ contradiction.
A: As an alternative, we can treat this as a simplification problem:$%
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\subcalch}[1]{\\ \quad & \quad #1 \\ \quad &}
\newcommand{\subcalc}{\quad \begin{aligned} \quad & \\ \bullet \quad & }
\newcommand{\endsubcalc}{\end{aligned} \\ \\ \cdot \quad &}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
%$
$$\calc
    \langle \forall c :: c > 0 \;\then\; a < b+c \rangle
\op=\hint{arithmetic -- to isolate $\;c\;$ in right hand side}
    \langle \forall c :: c > 0 \;\then\; c > a-b \rangle
\op=\hint{logic: contraposition -- to make the next step more natural}
    \langle \forall c :: c \le a-b \;\then\; c \le 0 \rangle
\op=\hint{$\;\le\;$ is reflexive and transitive}
    a-b \le 0
\op=\hint{arithmetic}
    a \le b
\endcalc$$
