jmc

Let the two numbers be $a$ and $b.$ Then $\sqrt{ab}=\sqrt{3},$ so $ab=3.$ Also, $$\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}=\frac{3}{2},$$so $a+b=\frac{4}{3}ab=4.$

Then by Vieta's formulas, $a$ and $b$ are the roots of the quadratic $$x^2-4x+3=(x-1)(x-3),$$so the two numbers are $\boxed{1,3}.$