Best approximation and an inequality Let $H$ be a Hilbert space. Let $E\subset H$ and $x\notin E$. Suppose that there exists $y^*\in E$ such that $$\|x-y^*\|=\min_{y\in  E}\|x-y\|$$ (i.e., $y^*$ is the best approximant of $x$).
I hope (from geometric intuition) that the following inequality is true.
$$\|x'-x\|^2+\|x'-y'\|^2\ge \|x'-y^*\|^2$$ for every $x'\in H$ and $y'\in E$.
Can anyone prove this?
 A: Take
\begin{align}
H &= \mathbb{C} \\
E &= \lbrace 0, 4 \rbrace \\
x &= 1 \\
x' &= 3 \\
y' &= 4
\end{align}
Then $y^* = 0$, and the conjecture is proven false.
A: I assume $E$ is convex and $H$ is a real Hilbert space.
Wlog, $y^\ast = 0$.  Since $E$ is convex and $0$ is the closest point in it to $x$, we have $\langle y,x\rangle \le 0$ for all $y\in E$.
If $\langle x,x'\rangle \le 0$ then
$$
\|x'-x\|^2 + \|x'-y'\|^2
\ge \|x'-x\|^2
= \|x'\|^2 + \|x\|^2 - 2\langle x,x'\rangle
\ge \|x'\|^2
$$
as desired.
If $\langle x,x'\rangle > 0$ then decompose $x' = \lambda x + v$, where $\lambda > 0$ and $\langle x,v\rangle = 0$.  Note that $v$ is the nearest point in $x^\perp$ to $x'$, and that since $\langle x,x'\rangle > 0$, all points in $E$ are on the other side of $x^\perp$ from $x'$, so they are further away from $x'$ than $v$; so $\|x'-y'\|\ge\|x'-v\|$.  Also, $\lambda x$ is the nearest point in the span of $x$ to $x'$; so $\|x'-x\|\ge\|x'-\lambda x\|$.  Thus
$$
\|x'-x\|^2 + \|x'-y'\|^2
\ge \|x'-\lambda x\|^2 + \|x'-v\|^2
= \|x'\|^2
$$
again as desired.
(Incidentally, the problem is really two-dimensional; everything of interest happens in the plane through $x$, $x'$, and $y^\ast$.  So if you want, you can draw a picture and do everything with classical Euclidean geometry.)
