I have a problem with this proof, I don't understand one equality.
REMARK: Let $R$ be a ring, let $I,J \triangleleft R$ then $\mathrm{rad}(I+J) = \mathrm{rad}(\mathrm{rad}(I) + \mathrm{rad}(J))$.
PROOF: Notice that $I+J = (I \cup J) \subseteq (\mathrm{rad}(I)\cup \mathrm{rad}(J)) = \mathrm{rad}(I) + \mathrm{rad}(J)$. Fix $a \in \mathrm{rad}(\mathrm{rad}(I) + \mathrm{rad}(J))$. Then $a^n \in \mathrm{rad}(I) + \mathrm{rad}(J)$ for some $n \in \mathbb{N}$. Hence $a^n = b+c$ with $b\in \mathrm{rad}(I)$ and $c \in \mathrm{rad}(J)$, that is $b^k \in I$ and $c^l \in J$ for some $k,l \in \mathbb{N}$. Therefore $a^{n(k+l)} = (a^n)^{k+l} = (b+c)^{k+l} = b^k x+ c^{l} y$ for some $x,y \in R$, as a result, $a\in \mathrm{rad}(I+J)$.
My equation is: Why $(b+c)^{k+l} = b^k x+ c^{l} y$ for some $x,y \in R$ ??
Note
$$(b+c)^{k+l} = \sum_{i=0}^{k+l} \binom{k+l}{i}b^{k+l-i}c^i = \sum_{i=0}^{l} \binom{k+l}{i}b^{k+l-i}c^i+\sum_{i=l+1}^{k+l} \binom{k+l}{i}b^{k+l-i}c^i = \\ = b^k \sum_{i=0}^{l} \binom{k+l}{i}b^{l-i}c^i+c^l\sum_{i=l+1}^{k+l} \binom{k+l}{i}b^{k+l-i}c^{i-l},$$
so by choosing
$$x:=\sum_{i=0}^{l} \binom{k+l}{i}b^{l-i}c^i$$
and
$$y:=\sum_{i=l+1}^{k+l} \binom{k+l}{i}b^{k+l-i}c^{i-l}$$
we conclude $(a+b)^{k+l}=b^kx+c^ly$.
Addendum. In any case, there is a cleaner way of proving your remark:
$I \subseteq rad(I)$ and $J \subseteq rad(J)$, so $I+J \subseteq rad(I)+rad(J)$. Hence $rad(I+J) \subseteq rad(rad(I)+rad(J))$.
Conversely, $I \subseteq I+J$, so $rad(I) \subseteq rad(I+J)$. Similarly, $rad(J) \subseteq rad(I+J)$. Hence $rad(I)+rad(J) \subseteq rad(I+J)$, so $rad(rad(I)+rad(J)) \subseteq rad(I+J)$.