Relationship between two projectors Please, somebody can help me with this problem?

Let $V$ and $W$ be two closed subspaces of a Hilbert $(H, \langle \cdot,\cdot\rangle)$, and let $P:H\rightarrow V$ and $Q:H\rightarrow W$ the orthogonal projectors respectively. Show that
$$\langle\ (Q-P)(x),\ x\ \rangle\ \geq\ 0,\ \forall\ x\in H\quad \mbox{ if and only if }\quad V\subseteq W.$$


Thanks in advance.
 A: Suppose first that $V\not\subseteq W$. Then there exists $v\in V\setminus W$. Let $x=v-Qv$. Then $Qx=0$ $Px\ne0$, and we have
$$
\langle (Q-P)x,x\rangle=\langle -Px,x\rangle=-\langle Px,x\rangle=-\langle Px,Px\rangle=-\|Px\|^2<0.
$$
As this is a contradiction, we have $V\subseteq W$.
Conversely, if $V\subseteq W$, then for any $x\in H$ we have $QPx=Px$ (as $Px\in V$), so $QP=P$. Now $P,Q$ are selfadjoint, so taking adjoints we get $PQ=P$. Then
$$
(Q-P)^2=Q+P-PQ-QP=Q-P.
$$
Then, for any $x\in H$,
$$
\langle (Q-P)x,x\rangle=\langle (Q-P)^2x,x\rangle=\langle (Q-P)x,(Q-P)x\rangle=\|(Q-P)x\|^2\geq0.
$$
A: 1) Preliminary facts: for a projection $P$ (i.e. a self-adjoint idempotent $P$), we have the following characterization of the range of $P$:

$$
\|Px\|\leq \|x\|\quad\forall x\in H\quad\mbox{and}\quad \mbox{im}P=\{x\,;\,\|Px\|=\|x\|\}.
$$

Recall that like for any idempotent, $\mbox{im} P=\{x\,;\,Px=x\}=\ker (I-P)$ and $\ker P=\mbox{im}(I-P)$. 
Take this opportunity to observe that for two idempotents: 
$$\mbox{im} P\subseteq \mbox{im} Q \iff \mbox{im}P\subseteq \ker (I-Q) \iff (I-Q)P=0\iff P=QP.$$
Also $H=\ker P\oplus \mbox{im}P$, where the decomposition of $x\in H$ according to this direct sum is $x=Px+(I-P)x$. When $P$ is self-adjoint, this sum is orthogonal, and conversely. Therefore $\|x\|^2=\|Px\|^2+\|(I-P)x\|^2$. It is now clear that $\|Px\|\leq \|x\|$ in general, and that $Px=x$ iff $(I-P)x=0$ iff $\|x\|=\|Px\|$.
Also recall, that when $P$ is a projection, $(Px,x)=(P^2x,x)=(Px,P^*x)=(Px,Px)=\|Px\|^2$.
2) Proof: so for two projections $P,Q$
$$
((Q-P)x,x)\geq 0 \iff (Qx,x)\geq (Px,x)\iff \|Qx\|\geq \|Px\|.
$$
If the latter holds for every $x\in H$, we see that $\|x\|\geq \|Qx\|\geq \|Px\|=\|x\|$ implies $\|x\|=\|Qx\|$, that is $\mbox{im P}\subseteq \mbox{im } Q$. Conversely, $\mbox{im P}\subseteq \mbox{im } Q$ means $QP=P$.  Whence $PQ=(QP)^*=P^*=P$. So for every $x\in H$, we have $\|Px\|=\|PQx\|\leq \|Qx\|$ which finishes the proof.
