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 point $O$ onto sides $B C$, $C A$, $A B$ are points $D$, $E$, $F$ respectively, find the maximum value of $O D+O E+O F$.
One, Solution: As shown in the figure, since $O D \perp B C, O E \perp A C$, therefore, $D, E$ are the midpoints of $B C, A C$, respectively, $D E=\frac{1}{2} A B$, and $O, D, C, E$ are concyclic. By Ptolemy's theorem, we have $O D \cdot \frac{A C}{2} + O E \cdot \frac{B C}{2} = O C \cdot \frac{A B}{2}$.
That is, $O D \cdot A C + O E \cdot B C = O C \cdot A B$.
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 $O D(A B + A C) + O E(B C + A B) + O F(A C + B C) = R(A B + B C + C A)$.
Let the inradius of the triangle be $r$, then $O D \cdot B C + O E \cdot A C + O F \cdot A B = 2 S_{\triangle A B C} = (A B + B C + C A) r$. Adding the above two equations gives $(O D + O E + O F)(A B + B C + C A) = (R + r)(A B + B C + C A)$.

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