Hey I just found a solution on my own.
Let $a$ be an arbitrary Element of a Hilbert space.
We have
$\Vert P_2 P_1 a\Vert^2=\langle P_2 P_1a,P_1a\rangle=\Vert P_2 P_1 a\Vert\Vert P_1a\Vert\langle \frac{P_2P_1a}{\Vert P_2 P_1 a\Vert},\frac{P_1a}{\Vert P_2 P_1 a\Vert}\rangle<\epsilon\Vert P_2 P_1 a\Vert\Vert P_1a\Vert$$\Vert P_2 P_1 a\Vert^2=\langle P_2 P_1a,P_1a\rangle=\Vert P_2 P_1 a\Vert\Vert P_1a\Vert\langle \frac{P_2P_1a}{\Vert P_2 P_1 a\Vert},\frac{P_1a}{\Vert P_1 a\Vert}\rangle<\epsilon\Vert P_2 P_1 a\Vert\Vert P_1a\Vert$.
$\implies \Vert P_2 P_1 a\Vert<\epsilon\Vert P_1a\Vert$
$\implies\Vert P_2 P_1 a\Vert^2<\epsilon^2\Vert P_1a\Vert^2$
$\implies\langle P_1P_2P_1a,a\rangle\le\epsilon^2\langle P_1a,a\rangle$.
Since
$-\epsilon\Vert P_1a\Vert^2<-\Vert P_2 P_1a\Vert^2<\Vert P_2 P_1a\Vert^2$,
we also have
$-\epsilon^2\langle P_1a,a\rangle\le\langle P_2P_1P_2a,a\rangle$$-\epsilon^2\langle P_1a,a\rangle\le\langle P_1P_2P_1a,a\rangle$.
Changing the roles of $P_1$ and $P_2$ yields the claim.
