Let $a$ and $b$ be the positive roots of $x^2 - 3x + 1 = 0.$  Find
\[\frac{a}{\sqrt{b}} + \frac{b}{\sqrt{a}}.\]
By Vieta's formulas, $a + b = 3$ and $ab = 1.$

Let
\[t = \frac{a}{\sqrt{b}} + \frac{b}{\sqrt{a}}.\]Then
\begin{align*}
t^2 &= \frac{a^2}{b} + 2 \sqrt{ab} + \frac{b^2}{a} \\
&= \frac{a^3 + b^3}{ab} + 2 \\
&= \frac{(a + b)(a^2 - ab + b^2)}{ab} + 2 \\
&= \frac{(a + b)((a + b)^2 - 3ab)}{ab} + 2 \\
&= \frac{3 \cdot (3^2 - 3)}{1} + 2 \\
&= 20,
\end{align*}so $t = \sqrt{20} = \boxed{2 \sqrt{5}}.$