Given $C\subset\mathbb R^n$ where $C$ is convex and closed, let $f(a)=argmin_{x\in C}||x-a||^2$. I'd like to show $\langle f(x)-f(y),y-f(y)\rangle\le 0$. It's clear in $\mathbb R^2$ why this is true: if the angle between $f(b)-f(a)$ and $a-f(a)$ is smaller than 90 degrees (i.e. the inner product is positive), then $f(a)$ is further from $a$ than the point on the line between $f(a)$ and $f(b)$ that makes a right triangle with $a$ and $f(a)$ - this point is also in $C$ by convexity, contradicting that $f(a)$ is the closest point in $C$ to $a$.
But I'm not sure how to show this in $\mathbb R^n$. Any pointers greatly appreciated.
Hint: for any $x ,y \in \Bbb R^n$ we have $f(x) , f(y) \in C$ and $ \| f(y) -y \| = d (y ,C) $ Hence $$\| f(y) - y \| \leq \| y - f(x) \| $$ so $$\| f(y) - y \|^2 \leq \| y - f(x) \|^2 $$
Then expand both sides to see what happens .