Distance from a convex set to a point Let $Y \in \mathbb{R}^n$ be a nonempty convex set  such that $0 \notin Y$ and fix $y_1,\dots,y_n$ in $Y$, where $n \ge 2$. I know that there exist $i,j$ such that $\Vert y_i \Vert > \Vert y_j\Vert$. Define $ C = C(y_1,\dots,y_n)$, i.e. the set of convex combinations. Moreover, I know that there exists a unique $x\in C$ such that \begin{align*} \Vert x \Vert = d(0,C),\end{align*} where the latter is the distance from $0$ to $C$.
Now I want to show that \begin{align*}  \Vert x \Vert < \Vert (1-\lambda) x + \lambda y_l\Vert\end{align*} for all $\lambda \in (0,1)$, where $y_l$ such that $\Vert y_l \Vert = \max\{\Vert y_1 \Vert, \dots, \Vert y_n \Vert \}$.
Notice that the proof should be constructive, i.e. no law of excluded middle should be used. 
I did already show that $\langle x , c \rangle \ge 0$ for all $ c \in C$.
So far I was only able to show 
\begin{align*}
(1-\lambda)\lambda\Vert x \Vert < \Vert (1-\lambda) x + \lambda y_l\Vert.
\end{align*}

If I am understanding correctly, this seems quite trivial if the aformentioned distance function, $d$, computes the distance between the origin and the cet $C$ by finding a point $x$ in $C$ with minimal $L_2$-norm, i.e., $$ d(\overset{\to}{0}, C) = \min_{x \in C} \left| \left| x\right| \right|_2.$$
If that's the case, then it's easy to see that for any $i \in [n]$, and $\lambda \in [0,1]$, the following holds:
$$ \left| \left| x\right| \right|_2 < \left| \left| \lambda x + (1-\lambda) y_i\right| \right|_2.$$
The reason of why such inequality holds is by the fact that $x$ minimizes the distance to the origin (having the least norm, and such point being unique as assumed). Also note that each point $w$ displayed as $\lambda x + (1-\lambda) y_i$, by convexity of $C$, it holds that $w \in C$, thus $\left| \left| w\right| \right|_2 > \left| \left| x\right| \right|_2$.
