Proof about inequalities Let $a,b,c \in \mathbb{R}$. Prove that if for $a<c$ $\forall$ $c>b$, then $a \leq b$. 
My attempted proof was that if we take the contrapositive of the statement, then we get the inequality $c \leq b$ $\forall$ $a \geq c$. Then if we add the the given inequality to this we get $a+c \leq b+c$ which proves that $a \leq b$. I am almost certain that there is something wrong with this approach. Any hints?

To take the contrapositive of a statement you need to be extra careful about quantifiers.
The contrapositive of $$"a<c \text{ holding for all } c>b \text{ implies } a \le b."$$ is instead 
$$"a > b \text{ implies that there exists } c>b \text{ such that }a \ge c."$$  Note that the 'for all $c>b$' turned into a 'there exists $c>b$' in the process of negation.  
