Theorem:  Prove that $J_{\ell}(R)=\{x \in R \mid 1-y x$ is a unit, $\forall \; y \in R\}$.

Proof: $(\Rightarrow)$ : 

Assume $x \in J_{\ell}(R)$.

This means $x$ lies in every maximal left ideal of $R$.

We want to show that for every $y \in R$, the element $1 - yx$ is a unit.

Suppose, for contradiction, that $1 - yx$ is not a unit for some $y \in R$.

In particular, assume it has no left inverse.

Then there exists a maximal left ideal $M$ such that $1 - yx \in M$.

Since $x \in J_{\ell}(R)$, we also have $x \in M$, and because $M$ is a left ideal, $yx \in M$ as well. Then:

$$1 = (1 - yx) + yx \in M,$$
which contradicts the fact that $M$ is a proper ideal (it cannot contain $1$).

So $1 - yx$ must have a left inverse.

Suppose $z(1 - yx) = 1$ for some $z \in R$, but suppose $z$ is not a unit.

Then $z$ must lie in some maximal left ideal $M'$, and again since $x \in J_{\ell}(R)$, we have $x \in M'$.

Then:
$$1 = z(1 - yx) = z - zyx \in M',$$
which is again a contradiction.

Hence, $z$ must be a unit, and therefore so is $1 - yx$.

This proves the forward direction.

$(\Leftarrow)$ : Now assume $x \in R$ is such that $1 - yx$ is a unit for all $y \in R$.

We want to show $x \in J_{\ell}(R)$, meaning it lies in every maximal left ideal.

Suppose not — then there exists some maximal left ideal $M \subset R$ such that $x \notin M$.

Then $M + Rx = R$, which means there exist $z \in M$, $a \in R$ such that:
$$1 = z + ax \Rightarrow z = 1 - ax.$$
But $1 - ax$ is a unit by assumption, so $z \in M$ is a unit.

This is a contradiction, since maximal left ideals cannot contain units (otherwise they would be the whole ring). 

Thus, $x \in J_{\ell}(R)$.