Every radical ideal in a Noetherian ring is a finite intersection of primes

Prove that if $I$ is a radical ideal and $ab\in I$, then $I=\operatorname{rad}(I+(a))\cap \operatorname{rad}(I+(b))$.

Deduce that every radical ideal in a Noetherian ring is a finite intersection of primes.

I've done the first part (showing the sets are equal), but don't know how to do the second.

If $I$ is a radical ideal which is not prime, then there are elements $a,b$ with $ab \in I$ but $a \notin I$ and $b \notin I$. By the first part you can write $I$ as an intersection of two radical ideals $I_1, I_2$ containing $a$ and $b$ respectively. The same procedure can be applied for $I_1$ and $I_2$, and since the ring is Noetherian you will reach prime ideals after finitely many steps (otherwise there would be an infinite ascending chain of (radical) ideals).