Suppose that C is a closed convex subset of $\mathbb R^n$ and $x \in \mathbb R^n$. The projection of $\mathbf x$ onto C is the closest point $\mathbf y \in C : \mathbf z = \mathbf y$ minimizes ||$\mathbf z - \mathbf x$||$_2$ over all $\mathbf z \in C$.
If $P_C$($\mathbf x$)=$\mathbf y$ is the projection function, show that $(\mathbf x - \mathbf y)^T (\mathbf z- \mathbf y) \leq 0$ for all $\mathbf z \in C$.
Hint: look at the derivative of ||$(\mathbf x - (\mathbf y + \theta (\mathbf z - \mathbf y))$||$_2^2$ with respect to $\theta$ at $\theta = 0$.
I also went to my professor's office hours and he helped me get started in the following way but I don't know where to go from here...
We have $$||\mathbf x - \theta(\mathbf z - \mathbf y)|| \geq ||\mathbf x - \mathbf y||$$ and by squaring both sides we get $$||\mathbf x - \theta(\mathbf z - \mathbf y)||^2 - ||\mathbf x - \mathbf y||^2 \geq 0$$ and then we can take the limit: $$\lim_{\theta\to0^+}\frac{||\mathbf x - \theta(\mathbf z - \mathbf y)||^2 - ||\mathbf x - \mathbf y||^2}{\theta} $$
This is an extension of a question that I already asked Minimize Function over Convex Subset
The equation you want to use is this: $$||\mathbf x -\mathbf y - \theta(\mathbf z - \mathbf y)||^2 - ||\mathbf x - \mathbf y||^2 \geq 0$$ Remember the reason this is true is that $\mathbf y + \theta(\mathbf z - \mathbf y) \in C$ for all $\theta \in [0,1]$.
To make things a little easier, first show that if $$||\mathbf a - \theta \mathbf b||^2 - ||\mathbf a||^2 \geq 0$$ for all $\theta \in [0,1]$, then $\mathbf a^T \mathbf b \le 0$.