This comes from Wasserman's "Exact C*-Algebras and Related Topics", lemma 3.10.
If $G$ is a group that has Propety T and generating set $1=g_1,...,g_n$, then there exists a $C<n$ such that if $\pi$ and $\rho$ are unitary representations of G onto Hilbert spaces $H_\pi$ and $H_\rho$ such that $||\sum_{i=1}^{n}\pi(g_i)\otimes\rho(g_i)||_{min}\geq C$, then there is a finite dimensional unitary representation $\pi'$ that is contained in both $\pi$ and $\rho$.
The proof shows a particular inequality must hold, and then you can choose C large enough so that the last inequality below is less than the Kazhdan constant for the generating set of the group. This is the beginning of the proof:
Since $||\sum_{i=1}^{n}\pi(g_i)\otimes\rho(g_i)||_{min}\geq C$, there exists a $\xi\in H_\pi\otimes H_\rho$ such that $||\sum_{i=1}^{n}(\pi(g_i)\otimes\rho(g_i))\xi||\geq C$. Now $1=g_1$, so that $(\pi(1)\otimes\rho(1))\xi=\xi$, and hence for $i\geq2$, we have that $||(\pi(g_i)\otimes\rho(g_i))\xi-\xi||\leq2\sqrt{n-C}$.
I just cannot seem to figure out why this last inequality is true. I've done it when $n=2$ using the parallelogram identity, but it doesn't seem to generalize to anything larger than $n=2$. I haven't used anything about the fact that we're working with tensors of representations, but for some reason I'm convinced that it isn't relative to why the equality holds, but perhaps I'm mistaken.
So it turns out the n=2 case was very helpful. I forgot to mention that $\xi$ is a unit vector.
To reduce clutter, set $T_i=\pi(g_i)\otimes\rho(g_i)$. We suppose that $C\leq||\sum_{i=1}^{n}T_i\xi||\leq n$. Hence for any $i\geq 2$, we have that $C-(n-2)=C-n+2\leq||T_i\xi+\xi||$ by the reverse triangle inequality. Using the parallelogram identity, we have that
\begin{align} ||T_i\xi-\xi||^2 &= 2||\xi||^2+2||T_i\xi||^2-||T_i\xi+\xi||^2 \\ & \leq 4-(C-n+2)^2 \\ &=(4+C-n)(n-C)\\ &<4(n-C), \end{align}
where the first $\leq$ comes from the above paragraph, the line after that is the difference of two squares after simplifying, and the last holds from the fact that $C<n$, i.e. $C-n<0$.