Prove that $ I_{R} (A) $ is the "biggest" subring in $R$ in which $A$ is ideal. 
Let $A$ be subring of ring $R$. Let $ I_{R} (A) = \{ x \in R \colon xa, ax \in A, \forall a \in A \} $. Prove that $ I_{R} (A) $ is the maximal (biggest) subring of $R$ in which $A$ is ideal.

I easily proved that $I_{R} (A) $ is subring of $R$, and that $A$ is ideal of $I_{R} (A) $. I am having trouble with proving that $ I_{R} (A) $ is the maximal subring. I tried to prove it directly, i.e. observe will every subring of $R$ in which $A$ is ideal be subset of $I_{R} (A)$, but since I am still beginner in this scope I got stuck. Any hint helps.

Let $A\subseteq S\subseteq R$ with $S$ a ring and suppose $A$ is an ideal of $S$. If $x\in S$, then $xa$, $ax\in A$ for all $a\in A$. Therefore $x\in I_R(A)$. So $S\subseteq
I_R(A)$.
