Ideal products and calculating with ideals

Let $R$ be a ring and $I,J$ ideals of $R$. Let $$IJ = \left\{ \sum_{i=1}^n a_ib_i \big\vert\ n \geq 0, a_1,\cdots, a_n \in I, b_1,\cdots,b_n \in J \right\}$$ i) Show that IJ is an ideal of R and $IJ \subset I \cap J$.

ii) Let $K$ be another ideal of $R$. Show that $I(J + K) = IJ + IK$.

i)

In order to show that $IJ$ is an ideal, I need to show that $\forall r \in R\ \forall s \in IJ: r s r^{-1} \in IJ$, right?

Let $r \in R$ be arbitrary, $s\in IJ = (a_1b_1)+\cdots+(a_kb_k)$. $rsr^{-1} = r( (a_1b_1)+\cdots+(a_kb_k))r^{-1} = (\underbrace{ra_1}_{\in I}b_1)+\cdots+(\underbrace{ra_k}_{\in I}b_k) \in IJ$, as I is an ideal of R, and therefore any $r\cdot a_i$ is, right?

So what about $IJ \subset I \cap J$? If I consider a single summand $a_ib_k$, that the result need to be in I as well as in J, because both are ideals of R. But why does this apply for the whole sum?

ii)

I have no idea, do you have a hint for me?

(1). Let $s = \sum a_ib_i \in IJ, r \in R.$ We need to show that $rs, sr \in IJ.$ Now $rs = \sum (ra_i)b_i \in IJ$ (since $I$ is an ideal) and $sr = \sum a_i(b_ir) \in IJ$ (since $J$ is an ideal).

To show that $IJ \subseteq I \cap J,$ first note that for any $a \in I, b\in J$ we have $ab \in I$ and $ab \in J$ (using the fact that both $I, J$ are ideals) and hence $ab \in I \cap J.$ Since $I \cap J$ is an ideal, $\sum a_ib_i \in I \cap J, \forall a_i \in I, b_i \in J$ ("sum" is a finite sum).

(2). $s = \sum a_i(b_i+c_i) \in I(J+K) \Rightarrow s = \sum (a_ib_i + a_ic_i) = \sum a_ib_i + \sum a_ic_i \in IJ + IK.$ For the reverse inclusion, first note that $J \subseteq J+K, K \subseteq J +K.$ So we have $IJ \subseteq I(J+K), IK \subseteq I(J+K)$ and hence $IJ + IK \subseteq I(J+K).$