For real numbers $t > 3,$ find the minimum value of
\[\frac{t}{\sqrt{t - 3}}.\]
Let $a = \sqrt{t - 3}.$  Then $a^2 = t - 3,$ so $t = a^2 + 3.$  Then
\[\frac{t}{\sqrt{t - 3}} = \frac{a^2 + 3}{a} = a + \frac{3}{a}.\]By AM-GM,
\[a + \frac{3}{a} \ge 2 \sqrt{3}.\]Equality occurs when $a = \sqrt{3},$ or $t = 6,$ so the minimum value is $\boxed{2 \sqrt{3}}.$