I was given an example to take the ideal of the irreducible polynomial $(x^2 + p),$ where $p$ is a prime and the ideal $(p)$ to show that the radical of the sum of ideals is not equal to the sum of the radicals.
But I do not quite understand the role of $x$ here and how it shows that the radical of the sum of ideals is not equal to the sum of the radicals. I also found this question here on the site Radical ideals in $\mathbb Z[x]$ such that their sum is not radical but still I do not understand.
We work in the ring $\mathbb{Z}[X]$. Any irreducible polynomial generates a prime ideal, which is thus a radical ideal. In particular, if $p$ is a prime number, then $I := \langle X^2 + p \rangle$ and $J := \langle p \rangle$ are radical ideals in $\mathbb{Z}[X]$. Their sum is $I+J=\langle X^2 + p , p \rangle = \langle X^2 , p \rangle$. This is not a radical ideal, since $X^2 \in \langle X^2 , p \rangle$ but (you have to prove this) $X \notin \langle X^2 , p \rangle$. So in this case, $r(I)+r(J)$ is strictly contained in $r(I+J)$: this is the inclusion $\langle X^2 , p \rangle \subset \langle X , p \rangle$.