# An optimization problem

I need to prove the following result:

Given a real sequence $a=(a_n)_{n\in\mathbb{Z}}$ and a number $A>0$ then

$||a||_{1}\leq A$ if and only if there exists $b_n$ such that $-b_n\leq a_n\leq b_n$ holds for any n and $\sum\limits_{n}b_n\leq A$, moreover, the minimun of $\sum\limits_{n}b_n$ over all such $b_n$ is equal to $||a||_{1}$, where $||a||_{1}$ is the $l^{1}$ norm of the sequence $a=(a_n)_{n\in\mathbb{Z}}$

-

One way is easy: If $\|a\|_{1}\leq A$ then let $b_i=|a_i|$.
The other way is not much harder - we know $0\leq |a_i|\leq b_i$. Show this implies $$\|a\|_1=\sum_{i=1}^\infty |a_i|\leq \sum_{i=1}^\infty b_i\leq A$$