The answer to the first question is positive, and the lower bound is achieved, since the set of all probability measures on $[0,1]^k$ with uniform marginals is compact and since the integral of the bounded continuous function $(x_1,\ldots,x_k) \mapsto x_1\cdots x_k$ on $[0,1]^d$ depends continuously of the probability measure.

Yet, finding the minimum is not obvious. For all $i<j$, the conditional distribution of $(X_i,X_j)$ given $X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_{j-1},X_{j+1},\ldots,X_k$ should be the decreasing coupling.

The answer to the first question is positive, and the lower bound is achieved, since the set of all probability measures on $[0,1]^k$ with uniform marginals is compact and since the integral of the bounded continuous function $(x_1,\ldots,x_k) \mapsto x_1 \cdots x_k$ on $[0,1]^k$ depends continuously of the probability measure.

Moreover, given such a probability measure $\pi$ on $[0,1]^k$, the integral of $x_1 \cdots x_k$ with regard to $\pi$ is strictly positive since $x_1 \cdots x_k>0$ for $\pi$-almost every $(x_1,\ldots,x_k)$.

Yet, finding the minimum is not obvious. For all $i<j$, the conditional distribution of $(X_i,X_j)$ given $X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_{j-1},X_{j+1},\ldots,X_k$ should be the decreasing coupling.

The answer to the first question is positive, and the lower bound is achieved, since the set of all probability measures on $[0,1]^k$ with uniform marginals is compact and since the integral of the bounded continuous function $(x_1,\ldots,x_k) \mapsto x_1\cdots x_k$ on $[0,1]^d$ depends continuously of the probability measure.

Yet, finding the minimum is not obvious. For $k=3$, I intuitively expect that taking $X_2=1-X_1$ and $X_3$ equal to a decreasing function of $X_1(1-X_1)$ will give the minumum.

The answer to the first question is positive, and the lower bound is achieved, since the set of all probability measures on $[0,1]^k$ with uniform marginals is compact and since the integral of the bounded continuous function $(x_1,\ldots,x_k) \mapsto x_1\cdots x_k$ on $[0,1]^d$ depends continuously of the probability measure.

Yet, finding the minimum is not obvious. For all $i<j$, the conditional distribution of $(X_i,X_j)$ given $X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_{j-1},X_{j+1},\ldots,X_k$ should be the decreasing coupling.