jmc

By Vieta's Formulas, we have $a+b+c=9$, $ab+ac+bc=11$, and $abc=1$, so $\sqrt{abc}=1$. (Note that the roots $a$, $b$, and $c$ are positive.)

We have $$s^2=a+b+c+2\sqrt{ab}+2\sqrt{ac}+2\sqrt{bc}=9+2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}),$$so $s^2-9=2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})$. Squaring, we get $$\begin{array}{rl}{s}^4-18s^2+81& =4(ab+ac+bc+2\sqrt{ab}\sqrt{ac}+2\sqrt{ab}\sqrt{bc}+2\sqrt{ac}\sqrt{bc})\\ & =4[ab+ac+bc+2\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})]=4(11+2s)=44+8s,\end{array}$$so $s^4-18s^2-8s+37=0$. Therefore, $s^4-18s^2-8s=\boxed{-37}$.