Proving inequality too simple? If I would like to show that $\exists y \in R, \ \forall x \in R:(y-x)^2-6(y-x)>9$, does it make sense to say:
Consider $x \in R$. Let $y=x$ thus $y \in R$, and show that $9>0$? 
Or, would I need to find a more specific value of $y$? It just seems too straight forward.

You want to prove that there exists a $y$ such that for all $x$ it follows that $0<(y-x)^2-6(y-x)+9 = (y-x-3)^2.$ Note that this inequality holds if and only if $y-x\neq 3$.  However for any $y$, taking $x=  y-3$ yields precisely that $(y-x-3)^2=0$, contradicting your statement.  It seems what you have written is not true.
