Proving an inequality about orthogonal projections

Question

Let $P$ be an orthogonal projection onto a closed subspace of a Hilbert space $\mathcal{H}$, and let $u,v \in \mathcal{H}$. I want to prove

\begin{equation} \|u-v\|^2 \leq \|u-Pv\|^2 + \|u-Pu\|^2 + 2\|v-Pv\|^2. \end{equation}

I imagine this is really easy, but I can't spot how to do it. Any ideas?

Answer 1

Let $v_0 + v_1 := Pv + (I-P) v =v$ and $u_0 + u_1 := Pu + (I-P) u =u$ , then $$\Vert u - v \Vert ^2 = \Vert (u - v_0) - v_1 \Vert^2 = \Vert u - v_0\Vert^2- 2 \Re\langle u-v_0, v_1\rangle + \Vert v_1 \Vert^2 $$

But $v_0,u_0 \perp v_1 $, thus $\langle u-v_0, v_1\rangle = \langle u, v_1\rangle = \langle u_1, v_1\rangle$, hence $$ - 2 \Re\langle u-v_0, v_1\rangle \le 2 \vert \langle u_1, v_1\rangle \vert \le 2 \Vert u_1 \Vert\Vert v_1 \Vert \le \Vert u_1 \Vert^2 + \Vert v_1\Vert^2,$$ by Cauchy-Schwarz and Young's inequality.