Let $A$ be a positive definite matrix and $v \in \mathbb{R}^n$ be a given vector. Suppose there exists a $y \in \mathbb{R}^n$ such that $0 < y^{\top}Ay \leq 1 $ and the following holds: $$ \langle x, 2v-x \rangle \leq \langle y, 2v -y \rangle $$ for all $0 < x^{\top}Ax \leq 1 $, where $\langle \cdot, \cdot \rangle$ is the standard inner product in $\mathbb{R}^n$.
Would it be possible to show $yi(2v_i-y_i)\geq 0$ for each $i \in \{1, \dots n\}$?
No. Let $S$ be the convex set $\{x^TAx\le1:x\in\mathbb R^n\}$. Suppose that we pick an entrywise positive vector $y$ on the surface $\partial S$ of $S$. Then $Ay$ is normal to the ellipsoid $\partial S$ at $y$ and it is pointing outward from $S$. If $Ay$ happens to have a negative coordinate, then for some sufficiently large $t>0$, the vector $v=y+tAy$ will have a negative coordinate and $v\notin S$ (because $t>0$ and $S$ is convex). E.g. when $$ A=\frac{1}{5}\pmatrix{2&-1\\ -1&1},\quad y=\pmatrix{1\\ 3},\quad Ay=\frac{1}{5}\pmatrix{-1\\ 2}\quad\text{and}\quad v=y+10Ay=\pmatrix{-1\\ 7}, $$ one may readily verify that $y\in\partial S$ (as $y^TAy=1$), $v\not\in S$ (as $v^TAv=13>1$) and $v_1<0$.
Note that $y$ is the closest point in $S$ to $v$ because it is the foot of perpendicular dropped from $v\notin S$ to the convex set $S$. Therefore $\|x-v\|^2\ge\|y-v\|^2$ for all $x\in S$. Expand both sides and rearrange terms, we obtain $\langle x,2v-x\rangle\le\langle y,2v-y\rangle$ for all $x\in S$. However, as $y$ is entrywise positive but $v$ has a negative coordinate, $y_i(2v_i-y_i)$ must be negative for some $i$. (In the example above, we have $y_1(2v_1-y_1)=-3<0$).