It is clear that for a tree $T$ $\min_i f_i(T)=\min {(i,j)\in E(T): X_i+\frac {X_i}{X_i+X_j}}$, where $E(T)$ is the set of edges of the tree $T$. Let $X_1\le X_2\le\dots X_n$. Then $\min_i f_i(T)\le X_1+\frac {X_1}{X_1+X_2}$, and the maximum is reached when the nodes corresponding to 1th and the 2nd invectors are connected by an edge.
It seems the following.
It is clear that for a tree $T$ $\min_i f_i(T)=\min {(i,j)\in E(T): X_i+\frac {X_i}{X_i+X_j}}$, where $E(T)$ is the set of edges of the tree $T$. Let $X_1\le X_2\le\dots X_n$. Then $\min_i f_i(T)\le X_1+\frac {X_1}{X_1+X_2}$, and the maximum is reached when the nodes corresponding to 1th and the 2nd invectors are connected by an edge.