difference of projections is a positive operator

Let $p,q:H\to H$ bounded, linear operator on a Hilbert space $H$, such that $p^*=p=p^2$ and $q^*=q=q^2$ ($p^*$ is the adjoint of $p$, for q the same. YOu call $p$ and $q$ a projection). Let $q(H)\subseteq p(H)$. Why is $\langle (p-q)(h),h\rangle \ge 0$ for all $h\in H$?

Here (bounded linear) orthogonal projections on Hilbert spaces is a discussion that $p$ is the identity on $im(p)$, zero on $im(p)^\perp$ and $q$ is the identity on $im(q)$, zero on $im(q)^\perp$. If $im(q)\subseteq im(p)$ and if you take $h\in H$ such that $h=h_1+h_2$ with $h_1\in im(p)$ and $h_2\in im(p)^\perp$, I have to do distinctions of cases now, right?

I think the proof shouldn't be difficult, but I'm stuck here. Can you explain me how to continue the proof?

Let $T = p-q$. Note that $p(H)$ and $p(H)^\perp$ are invariant subspaces of $T$. Consider the restriction of $T$ to $p(H)$. We can write $$T\mid_{p(H)} = \operatorname{id} - q$$ It should be easy to show that this operator is positive (I'll leave it to you).
Of course, the restriction of $T$ to $p(H)^\perp$ is $0$, which is a positive operator.
Note that $p(H)$ is closed since $p(H) = \ker p - \operatorname{id}$, so that $H = p(H) \oplus p(H)^\perp$. Since $T$ is positive on invariant subspaces whose direct sum is $H$, $T$ is positive on $H$.
In fact, we could have proven that $T$ is additionally a projection onto $p(H) \cap q(H)^\perp$.
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