For a given $y \in \mathbb{R}^n$, let $x_*=\arg\min_{x \in C}\frac{1}{2}\|x-y\|^2$ where $C$ is a convex set.
Can we show the following is nonnegative?
$$
\langle x_*, 2y-x_* \rangle \geq 0.
$$
My thoughts
By writing $2y=y+y$ and adding and subtracting manipulation, the above can be written as the following:
$$
\langle x_*, 2y-x_* \rangle= -\|x_*-y\|^2+\|y\|^2.
$$
When $y \in C$ we are done because $x*=y$, but when $y \notin C$ what can we do?
I can intuitively see when $y \notin C$, $\|y\|$ is greater than $\|x_*-y\|$ but I cannot show it algebraically.
Edit
Assume $C$ is not singleton set and contains the zero vector.
Best Answer
For the original (unedited question), let $y = (0,0)$, $y \not\in C$, and use what you've already found to disprove the statement.
If $C$ contains the zero vector (that is, $0\in C$), then by the arg min condition, $\|x_*-y\|^2\leq\|0 - y\|^2=\|y\|^2$. Then $0 \leq -\|x_*-y\|^2 + \|y\|^2 = \langle x_*, 2y-x_* \rangle$.
This works even if $C$ is a singleton, though of course non-singletons work too.