Real analysis supremum proof 
Let $A$ be a non-empty bounded sub-set of $\mathbb{R}$. Let
  $B\subset\mathbb{R}$, given by $$B=\left\{\frac{a_1+2a_2}{2} \,\Bigg|\,a_1,a_2\in
 A\right\}$$ Express $\sup B$ in terms of $\sup A$.

My attempt:
Suppose $a_1,a_2\in A$ and $b\in B$. 
Then $a_1 \leq \sup (A)$ and $a_2\leq \sup (A)$
So $a_1 + 2a_2\leq 3sup (A)$.
This gives $\frac{a_1 + 2a_2}{2}\leq \frac{3\sup (A)}{2}$.
This means that $\frac{3\sup (A)}{2}$ is an upperbound and $\sup(B)\leq\frac{3\sup (A)}{2}$. 
Now let $\epsilon>0$. 
$a_1>\sup(A) - \frac{2\epsilon}{3}$
$a_2>\sup(A) - \frac{\epsilon}{3}$
This gives $\frac{a_1 + 2a_2}{2}> \frac{3\sup (A)}{2}-\epsilon$. 
So this means that $\frac{3\sup(A)}{2}\leq\sup(B)$
So $\sup(B)=\frac{3\sup(A)}{2}$.
Is this correct? And how can I improve my proofs?

There are Lemmas that would be helpful.
$\sup(A+B) = \sup(A) + \sup(B)$
$\sup(cA) = c\sup(A), c\geq 0$
where A, B are subset of R, $A + B = \{z = a + b| a \in A, b \in B\}$, and c is a non-negative real number
then your questions can be answered: $\sup((A + 2*A)/2) = 3\sup(A)/2$
