17. (10 points) Let $\triangle A B C$ have $\angle A$, $\angle B$, and $\angle C$ opposite sides $a$, $b$, and $c$ respectively, and $a \cos C+\frac{1}{2} c=b$.
(1) Find the size of $\angle A$;
(2) If $a=1$, find the maximum value of the inradius $R$ of $\triangle A B C$.
(1) By the sine rule, we have
$$
\sin A \cdot \cos C+\frac{1}{2} \sin C=\sin B \text {. }
$$

Since $\sin B=\sin (A+C)$, the original equation becomes
$$
\frac{1}{2} \sin C=\cos A \cdot \sin C \Rightarrow \cos A=\frac{1}{2} \text {. }
$$

Since $\angle A$ is an interior angle of the triangle, $\angle A=\frac{\pi}{3}$.
(2) Given $S_{\triangle A B C}=\frac{1}{2}(a+b+c) R$, and
$S_{\triangle A B C}=\frac{1}{2} b c \sin A=\frac{\sqrt{3}}{4} b c$,
we have $\frac{\sqrt{3}}{4} b c=\frac{1+b+c}{2} \cdot R$
$$
\Rightarrow R=\frac{\sqrt{3}}{2} \cdot \frac{b c}{1+b+c} \text {. }
$$

By the cosine rule, we get
$$
\begin{array}{l}
1=b^{2}+c^{2}-b c \\
\Rightarrow(b+c)^{2}=1+3 b c \\
\Rightarrow b c=\frac{(b+c)^{2}-1}{3} .
\end{array}
$$

Thus, $R=\frac{\sqrt{3}}{6}(b+c-1)$.
Since $b c \leqslant\left(\frac{b+c}{2}\right)^{2}$, we have
$$
\frac{(b+c)^{2}-1}{3} \leqslant\left(\frac{b+c}{2}\right)^{2} \text {. }
$$

Therefore, $0 \leqslant b+c \leqslant 2$.
Hence, $R_{\max }=\frac{\sqrt{3}}{6}$.
The above equation is achieved when and only when $b=c=1$, i.e., $\triangle A B C$ is an equilateral triangle.