Minimum Plus Maximum Exceeds One

Key idea. Reflecting every sample through $\tfrac12$ swaps the minimum and the maximum and turns $\{U+V>1\}$ into $\{U+V<1\}$.

Set $Y_i=1-X_i$. The $Y_i$ are again i.i.d. $\mathrm{Unif}(0,1)$, so $(Y_1,\dots,Y_n)$ has the same joint law as $(X_1,\dots,X_n)$.

The map $x\mapsto 1-x$ is decreasing. Hence

The pair $(\min_i Y_i,\max_i Y_i)$ has the same law as $(U,V)$. Therefore

Next we show that the boundary case has probability zero. Since $U$ and $V$ are both among the $X_i$, the event $\{U+V=1\}$ forces $X_i+X_j=1$ for some pair $i,j$, with $i=j$ allowed. There are only finitely many such pairs, and each event $\{X_i+X_j=1\}$ has probability $0$, because $X_i+X_j$ is a continuous random variable. So $\mathbb{P}[U+V=1]=0$.

The two remaining probabilities are equal and sum to $1$. Hence, for every $n\ge 1$,