In that notation, $I$ and $J$ are ideals of a commutative ring $A$ and $\sqrt{I}$ denotes the radical of the ideal $I$.
The first implication is obvious. In fact, since $I \subset \sqrt{I}$ and $J \subset \sqrt{J}$, we have $A = I + J \subset \sqrt{I} + \sqrt{J}$. This shows that $I + J = A \Rightarrow \sqrt{I} + \sqrt{J} = A $.
For the other side, I need to show that $ \sqrt{I} + \sqrt{J} = A \Rightarrow I + J = A $. I've already demonstrated that $\sqrt{\sqrt{I} + \sqrt{J}} = \sqrt{I + J}$, then I just have to prove that $\sqrt{I + J} = A \Rightarrow I + J = A$, but I don't know how to proceed from here.
Assume $\sqrt I + \sqrt J = A$, i.e. $x + y = 1$ for some $x, y$ where $x^m \in I$ and $y^n \in J$.
Therefore, $(x+y)^{m+n-1} = 1$.
Observe that in the binomial expansion of $(x+y)^{m+n-1}$, each term is $k x^r y^s$ for some $k \in \Bbb Z$ and $r+s = m+n-1$. If $r < m$ and $s < n$, then $r \le m - 1$ and $s \le n - 1$, so $r + s \le m + n - 2$, contradiction; therefore either $r \ge m$ or $s \ge n$. Therefore, the term either belongs to $I$ or belongs to $J$. Therefore, $1 = (x+y)^{m+n-1} \in I + J$.
This is basically the proof that the radical of an ideal is an ideal.