Smallest n for which 3 over n produces 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient

Theorem
Consider proper fractions of the form $\dfrac 3 n$ expressed in canonical form.

Let Fibonacci's Greedy Algorithm be used to generate a sequence $S$ of Egyptian fractions for $\dfrac 3 n$

The smallest $n$ for which $S$ consists of $3$ terms, where $2$ would be sufficient, is $25$.

Proof
We have that:

But then we have: