jmc

By Cauchy-Schwarz, $$(y^2+x^2)(3x^2+y^2)\ge (xy\sqrt{3}+xy{)}^2,$$so $$\frac{\sqrt{(x^2+y^2)(3x^2+y^2)}}{xy}\ge 1+\sqrt{3}.$$Equality occurs when $\frac{y^2}{3x^2}=\frac{x^2}{y^2},$ or $y=x\sqrt[4]{3},$ so the minimum value is $\boxed{1+\sqrt{3}}.$