# Real analysis supremum proof

Let A be a non-empty bounded sub-set of $\mathbb{R}$. Let
$B\subset\mathbb{R}$, given by:$B=\{\frac{a_1+2a_2}{2}|a_1,a_2\in A\}$. 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?