Three, let the three sides of $\triangle A B C$ be $a, b, c$, and all are numbers. If it satisfies $\angle A=3 \angle B$, try to find the minimum value of the perimeter and provide a proof.
From the Law of Sines, we have \(\frac{a}{b} = \frac{\sin 3B}{\sin B} = (2 \cos B)^2 - 1\), \(\frac{c}{b} = (2 \cos B)^3 - 4 \cos B\). Since \(2 \cos B = \frac{a^2 + c^2 - b^2}{ac} \in \mathbb{Q}\), there must exist positive integers \(p, q\) with \((p, q) = 1\) such that \(2 \cos B = \frac{p}{q}\). Therefore, \(\frac{a}{b} = \left(\frac{p}{q}\right)^2 - 1 \Leftrightarrow \frac{a}{p^2 - q^2} = \frac{b}{q^2}\); \(\frac{c}{b} = \left(\frac{p}{q}\right)^3 - 2 \cdot \frac{p}{q} \Leftrightarrow \frac{c}{p^3 - 2pq^2} = \frac{b}{q^3}\). Thus, \(\frac{a}{(p^2 - q^2)q} = \frac{b}{q^3} = \frac{c}{p^3 - 2pq^2} = \frac{e}{f}\), where \((e, f) = 1\). This gives \(a = \frac{e}{f}(p^2 q - q^3)\), \(b = \frac{e}{f} q^3\), \(c = \frac{e}{f}(p^3 - 2pq^2)\). To minimize the perimeter, \((a, b, c) = 1\). Since \((e, f) = 1\), it follows that \(f \mid q^3\) and \(f \mid p^3 - 2pq^2\).

We now prove that \(f = 1\). If \(f > 1\), then \(f\) must have a prime factor \(f' > 1 \Rightarrow f' \mid q^3, f' \mid p^3 - 2pq^2 \Rightarrow f' \mid q \Rightarrow f' \mid 2pq^2 \Rightarrow f' \mid p^3 \Rightarrow f' \mid p\). This contradicts \((p, q) = 1\). Hence, \(f = 1\). Since \((a, b, c) = 1\), it follows that \(e = 1\). Therefore, \(a = p^2 q - q^3\), \(b = q^3\), \(c = p^3 - 2pq^2\), with \((p, q) = 1\). On the other hand, from \(0^\circ < A + \mathcal{Q} = 4B < 80^\circ \Rightarrow 0^\circ < B < 45^\circ \Rightarrow \sqrt{2} < 2 \cos B < 2 \Rightarrow \sqrt{2} q < p < 2q\). It is easy to see that the smallest \(p, q\) satisfying this inequality are \(p = 3, q = 2\). At this point, \(c = p(p^2 - 2q^2)\) is minimized when \(p\) is smallest and \(p^2 - 2q^2 = 1\) is also minimized. Hence, \(S = a + b + c = p^2 q + p(p^2 - 2q^2)\). When \(p = 3, q = 2\), we have \(S_{\text{min}} = 18 + 3 = 21\).