set the lenght of three sides $BC=a,AC=b,AB=c$ ,so $c=2b$ ，set $P$ as semi-perimeter ,so $P=\frac{a+b+c}{2}=\frac{a+3b}{2}$ \begin{aligned} \large{By **Heron's formula** S_{△ABC}=\sqrt{P(P-a)(P-b)(P-c)}=1 \\ \implies \frac{a+3b}{2}\frac{3b-a}{2}\frac{a+b}{2}\frac{a-b}{2}=1 \\ \implies (9b^{2}-a^{2})(a^{2}-b^{2})=16 \\ \implies 16×9=(9b^{2}-a^{2})(9a^{2}-9b^{2}) \\ By *basic*inequality*: XY≤(\frac{X+Y}{2})^{2} \\ \implies 16×9=(9b^{2}-a^{2})(9a^{2}-9b^{2})≤(\frac{8a^{2}}{2})^{2} \\ \implies 4a^{2}≥12 \\ \implies a≥\sqrt{3}}\end{aligned}

As a result of which $\boxed{BC=a≥\sqrt{3}}$