Solution
= Solution
Put $w=x-y_0$. For any $y\in Y$ and any real $t$, minimality gives
$$
\|w-ty\|^2-\|w\|^2=-2t\operatorname{Re}\langle w,y\rangle+t^2\|y\|^2\geq0.
$$
The derivative at zero must vanish, so $\operatorname{Re}\langle w,y\rangle=0$. In a complex <Hilbert space>, apply the same argument to $iy$ to get the imaginary part zero as well. In the real case the first conclusion is already enough. Hence
$$
\boxed{x-y_0\in Y^\perp.}
$$