Let $n<d$ be positive integers, and let $0< \epsilon\leq 1-\frac{n}{d}$ be a real number. I would like a succinct description of the following quantity:


$$
f\_{\epsilon}(n,d):=\min\_{\substack{x\_1\geq \dots \geq x\_d\geq 0\\ x\_1+\dots+x\_n =1-\epsilon\\ x\_{n+1}+\dots+x\_d=\epsilon}} \left(\sum\_{1 \leq i\_1 < \dots < i\_{n+1} \leq d} x\_{i\_1}\cdots x\_{i\_{n+1}}\right).
$$
In the absence of such a description, I would be interested in good lower/upper bounds on this quantity.


(I wasn't sure what tags to use.)


The assumption $\epsilon\leq 1-\frac{n}{d}$ ensures that the function is defined.



Bound on sum of products of positive real numbers with prescribed subset sums
Here is a succinct description of $f\_{\epsilon}(n,d)$ (I only have time to sketch the argument-- hopefully it is right).


Let
$$
h(x\_1,\dots, x\_d)=\left(\sum\_{1 \leq i\_1 < \dots < i\_{n+1} \leq d} x\_{i\_1}\cdots x\_{i\_{n+1}}\right),
$$
and let $x \in \mathbb{R}^d$ be a feasible element of the minimization (i.e., $x\_1\geq \dots \geq x\_d\geq 0, x\_1+\dots+x\_n=1-\epsilon$, and $x\_{n+1}+\dots+x\_d=\epsilon$).


For any $i<j \in \{n+1,\dots, d\}$, let


$$g(t)=h(x\_1,\dots, x\_{i-1},x\_i+t,x\_{i+1},\dots,x\_{j-1},x\_j-t, x\_{j+1},\dots, x\_d).$$


It is straightforward to verify that $g(t)$ is a quadratic function with negative second derivative. So $g(t)$ is minimized at an extreme point. It follows that $h(x)$ is minimized when $x\_{r+1},\dots, x\_n$ are as ``unbalanced" as possible. This implies that the minimum is attained for some $x \in \mathbb{R}^d$ for which
$$
(x\_{n+1},\dots, x\_d)=\left(\frac{1-\epsilon}{n},\dots, \frac{1-\epsilon}{n},\epsilon-\frac{k(1-\epsilon)}{n},0,\dots,0\right),
$$
where $\frac{1-\epsilon}{n}$ is repeated $k$ times in the tuple, and
$$
k=\lfloor \frac{\epsilon n}{1-\epsilon} \rfloor.
$$
We can do the same trick for the first $n$ coordinates of $x$, and prove that the minimum is attained for $x \in \mathbb{R}^d$ that satisfies
$$
(x\_1,\dots, x\_n)=\begin{cases} (\frac{1-\epsilon}{n},\dots,\frac{1-\epsilon}{n}) ,& \epsilon \geq \frac{1-\epsilon}{n}\\
(1-n\epsilon,\epsilon,\dots, \epsilon) ,& \epsilon\leq \frac{1-\epsilon}{n}.
\end{cases}
$$


It follows that
$$
f\_\epsilon(n,d)=\begin{cases} \binom{n+k}{n+1} \left(\frac{1-\epsilon}{n}\right)^{n+1}+\binom{n+k}{n} \left(\frac{1-\epsilon}{n}\right)^n \left[\epsilon-k\left(\frac{1-\epsilon}{n}\right)\right] ,& \epsilon \geq \frac{1-\epsilon}{n}\\
(1-n\epsilon)\epsilon^n,& \epsilon\leq \frac{1-\epsilon}{n}.
\end{cases}
$$


