6. In a right triangular prism $A B C-A_{1} B_{1} C_{1}$, $A B=1, B C=C C_{1}=\sqrt{3}, \angle A B C=90^{\circ}$, point $P$ is a moving point on the plane $A B C$, then the minimum value of $A_{1} P+\frac{1}{2} P C$ is $\qquad$ .
In the plane $A_{1} P C$, construct the line $D C$ on the opposite side of line $P C$ such that $\angle D C P=30^{\circ}$. Construct $P D \perp D C$ intersecting $D C$ at point $D$, then $A_{1} P+\frac{1}{2} P C=A_{1} P+P D$.
When $A_{1}, P, D$ are collinear, i.e., draw the perpendicular line $A_{1} H$ from point $A_{1}$ to line $D C$, at this time $A_{1} P+\frac{1}{2} P C$ achieves its minimum value
$A_{1} H=A_{1} C \sin \left(\angle A_{1} C P+30^{\circ}\right)$. (The height on side $D C$)
By the principle of the smallest angle, when point $P$ is on line $A C$, $\angle A_{1} C P=\angle A_{1} C A$ is the smallest. Therefore, in $\triangle A_{1} C E$, let the height on side $C E$ be $h$, then $2 S_{\triangle A_{1} C E}=A_{1} E \cdot A C=C E \cdot h \Rightarrow h=\frac{A_{1} E \cdot A C}{C E}=\frac{\sqrt{3}}{2} \cdot \frac{5 \sqrt{3}}{3}=\frac{5}{2}$. In summary, the minimum value of $A_{1} P+\frac{1}{2} P C$ is $\frac{5}{2}$.