一、(50 points) Let $\triangle A B C$ be inscribed in the unit circle $\odot O$, and the center $O$ is inside $\triangle A B C$. If the projections of $O$ onto sides $B C$, $C A$, and $A B$ are points $D$, $E$, and $F$ respectively, find the maximum value of $O D+O E+O F$.
As shown in Figure 3, since $O D \perp B C$ and $O E \perp A C$, it follows that $D$ and $E$ are the midpoints of $B C$ and $A C$, respectively. Therefore, $D E = \frac{A B}{2}$, and points $O, D, C, E$ are concyclic.
By Ptolemy's theorem, we have
$$
O D \cdot \frac{A C}{2} + O E \cdot \frac{B C}{2} = O C \cdot \frac{A B}{2},
$$

which simplifies to $O D \cdot A C + O E \cdot B C = O C \cdot A B$.
Similarly, $O E \cdot A B + O F \cdot A C = O A \cdot B C$,
$$
O F \cdot B C + O D \cdot A B = O B \cdot A C.
$$

Noting that $O A = O B = O C = R$ (the circumradius of $\triangle A B C$), adding the above three equations yields
$$
\begin{array}{l}
O D(A B + A C) + O E(B C + A B) + O F(A C + B C) \\
= R(A B + B C + C A).
\end{array}
$$

Let the inradius of $\triangle A B C$ be $r$, then
$$
\begin{array}{l}
O D \cdot B C + O E \cdot A C + O F \cdot A B \\
= 2 S_{\triangle A B C} = (A B + B C + C A) r.
\end{array}
$$

Adding the above two equations, we get
$$
\begin{array}{l}
(O D + O E + O F)(A B + B C + C A) \\
= (R + r)(A B + B C + C A).
\end{array}
$$

Thus, $O D + O E + O F = R + r \leqslant R + \frac{R}{2} = \frac{3}{2}$. The equality holds when $\triangle A B C$ is an equilateral triangle.
Therefore, the maximum value of $O D + O E + O F$ is $\frac{3}{2}$.