Let $\displaystyle A$ be bounded above, then prove that $\displaystyle \sup(c+A)=c+\sup A,$ $\displaystyle c\in \mathbb R.$

Let $\displaystyle a\in A,$ since $\displaystyle A$ is bounded above, there's supremum, hence $\displaystyle a\le\sup A\implies a+c\le c+ \sup A$ then $\displaystyle \sup(a+c)\le c+\sup A,$ I'm on the right track? How do I finish?

I think I'm wrong.