5. The volume of tetrahedron $ABCD$ is $\frac{1}{6}$, and it satisfies $\angle ACB=45^{\circ}, AD+BC+\frac{AC}{\sqrt{2}}=3$, then $CD=$
5. $\sqrt{3}$ Detailed Explanation: $V_{D-A B C} \leqslant \frac{1}{3} S_{\triangle A B C} \cdot A D=\frac{1}{6} A D \cdot B C \cdot \frac{A C}{\sqrt{2}}$, and $V_{D-A B C}=\frac{1}{6}$, thus $A D \cdot B C \cdot \frac{A C}{\sqrt{2}} \geqslant$ 1. Therefore, $A D+B C+\frac{A C}{\sqrt{2}} \geqslant 3 \sqrt[3]{A D \cdot B C \cdot \frac{A C}{\sqrt{2}}} \geqslant 3$, equality holds when and only when $A D=B C=\frac{A C}{\sqrt{2}}=1$, at this time $A D=1, A D \perp$ plane $A B C$, hence $C D=\sqrt{3}$.