Let $x_1$, $x_2\dots$, $x_n\in[0,1]$, prove that $$\sum_{k=1}^n\prod_{i\ne k}x_i\ge2n-n^2+(n-1)\sum_{k=1}^nx_k.\tag1$$
There is an obvious proof using linearity in each variable, by which we may assume $$(x_1,x_2\dots,x_n)\in\{0,1\}^n,$$ and it is very easy.
But I want a better inequality proof. I tried to let $x_i=\frac1{y_i}$, then $$(1)\iff\frac{y_1+y_2+\cdots+y_n}{y_1y_2\cdots y_n}\ge2n-n^2+(n-1)\sum_{k=1}^n\frac1{y_k}.$$ Perhaps EV theorem helps, but there is a strange constraint $y_i\in[1,+\infty)$.
By the way, there are two equality conditions: $$(x_1,x_2\dots,x_n)=(1,1,1\dots,1)\text{ or }(0,1,1\dots,1).$$
Let $y_j=1-x_j$ for all $1\leq j\leq n$. Then $1\leq y_j\leq 0$ and hence, by Weierstrass inequality, we have,
$$\prod_{i\neq j}(1-y_j)\geq1-\sum_{i\neq j}y_j\\\iff \prod_{i\neq j}x_j\geq 1-\sum_{i\neq j}(1-x_j).$$ Summing all inequalities, we get,
$$\sum_{i=1}^n\prod_{i\neq j}x_j\geq n-n(n-1)+(n-1)\sum_{j=1}^nx_j=2n-n^2+(n-1)\sum_{j=1}^nx_j.$$