Orthogonal projection to closed, convex subset in a Hilbert space I don't understand one step in the proof of the following lemma (Projektionssatz):
Let $X$ a Hilbert space with scalar product $(\cdot)_X$ and let $A\subset X$ be convex and closed. Then there is a   unique map $P:X\rightarrow A$ that satisfies: $\|x-P(x)\| = \inf_{y\in A} \|x-      y\|$. This is equivalent to the following statement:
(1) For all $a\in A$ and fixed $x\in X$, $\mbox{Re}\bigl( x-P(x), a-P(x) \bigr)_X \le 0$.
I don't understand the following step in the proof that (1) implies the properties of $P$:
Let $a\in A$. Then
$\|x-P(x)\|^2 + 2\mbox{Re}\bigl( x-P(x), P(x)-a \bigr)_X + \|P(x)-a\|^2 \ge \|x-P(x)\|^2$.
I don't understand the "$\ge$". How do we get rid of the term $\|P(x)-a\|$ on the left hand side?
Thank you very much!

It is non-negative! Hence, just drop it.
