I have the set $C:=C(x_1,\dots,x_n) \in \mathbb{R}^n$ of convex combinations of $x_i$'s. I know that there exists an $x_i$ such that $\Vert x_i \Vert > 0$ and $ 0 \notin \mathring{C}(x_1,\dots,x_n) = \{\sum_{i=1}^{n}\lambda_i x_i : \lambda_i \in (0,1) \text{ and sum up to 1}\}$. I also know that there exists a unique $ y \in C$ such that $\Vert y \Vert = d(0,C)$ and the following holds: \begin{align*}\>\forall c \in \mathring{C} : \> \langle y , c \rangle > 0. \end{align*} Now I am supposed to show that $\langle y , x_i \rangle > 0$ holds.
Can somebody give me a hint for this task?
Recall Kolmogorov's inequality: given a point $x \in \mathbb{R}^n$ and a point $z \in C$, then $z$ is the projection of $x$ onto $C$ if and only if $\langle x - z, c - z \rangle \le 0$ for all $c \in C$. In our case, $x = 0$ and $z = y$, so $$\langle y, c - y \rangle \ge 0 \implies \langle y, c \rangle \ge \|y\|^2 > 0,$$ since $\langle c, y \rangle > 0$, and hence $y \neq 0$. This holds for all $c \in C$, not just $\mathring{C}$, including the points $x_1, \ldots, x_n$.