Properties of Tensor Products (Tsirelson’s inequality) I am attempting the proof below:
$$(Q \otimes S + R \otimes S + R \otimes T − Q \otimes T)^2 = 4I + [Q, R] \otimes [S, T].$$
I get to this step by expanding:
$$ST \otimes (R + Q)(R - Q) + ST \otimes (R + Q)(R - Q) + (R + Q)^2 + (R - Q)^2,$$
but I don't know how to simplify from here. Any tips or hints would be helpful.
The full problem is:

Suppose
  $Q = \vec q \cdot \vec \sigma$, $R = \vec r \cdot \vec \sigma$, $S = \vec s \cdot \vec \sigma$, $T = \vec t \cdot \vec \sigma$, where $\vec q, \vec r, \vec s, \vec t$ are real unit vectors
  in three dimensions. 
Show that
  $$
(Q ⊗ S + R ⊗ S + R ⊗ T − Q ⊗ T)^2 = 4I + [Q, R] ⊗ [S, T]
$$
Use this result to prove that
$$
Q ⊗ S + R ⊗ S + R ⊗ T − Q ⊗ T ≤ 2\sqrt{2}
$$
so the violation of the Bell inequality found in Equation [earlier equation] is the maximum
  possible in quantum mechanics.

 A: A bit late, but here we go. Starting with the problem statement of showing 
$$(Q \otimes S + R \otimes S  + R \otimes T - Q \otimes T)^2 = 4I + [Q, R] \otimes [S, T]$$
which we can re-write as the following:
$$[(Q+R) \otimes S + (R-Q)\otimes T]^2 = 4I + (QR - RQ) \otimes (ST - TS)$$
We can now expand the left side like so:
$$(Q+R)^2 \otimes S^2 + (R-Q)^2 \otimes T^2 + (R+Q)(R-Q)\otimes ST + (R-Q)(R+Q)\otimes TS$$
$$(Q+R)^2 \otimes S^2 + (R-Q)^2 \otimes T^2 + (R^2-RQ+QR-Q^2)\otimes ST + (R^2+RQ-QR-Q^2)\otimes TS$$
Taking out the $RQ$ and $QR$ terms from the last 2 sets of parenthesis in the expression above, we get:
$$(Q+R)^2 \otimes S^2 + (R-Q)^2 \otimes T^2 + (R^2-Q^2)\otimes ST + (R^2-Q^2)\otimes TS + (QR - RQ)\otimes ST - (QR-RQ)\otimes TS$$
which equals
$$(Q+R)^2 \otimes S^2 + (R-Q)^2 \otimes T^2 + (R^2-Q^2)\otimes ST + (R^2-Q^2)\otimes TS + (QR - RQ)\otimes (ST-TS)$$
But the last term in the expression above is also present on the right hand side of the equation in the problem statement, so this can be removed from both sides. So we have to prove:
$$(Q+R)^2 \otimes S^2 + (R-Q)^2 \otimes T^2 + (R^2-Q^2)\otimes ST + (R^2-Q^2)\otimes TS = 4I$$
The next step is to see that $Q^2, S^2, R^2, T^2= I$. There are many ways to show this, I will write one of the ways. Since all pauli matrices are hermitian we know that Q, R, S, and T are also hermitian. This means we can use the spectral decomposition theorem to show that these matrices are diagonizable. In exercise 2.60 of Nielsen & Chuang, we find that Q has eigenvalues +1, -1. Thus, 
$$Q = |q_1\rangle \langle q_1| - |q_2\rangle \langle q_2|$$
where $|q_1\rangle$ and $|q_2\rangle$ form an orthonormal basis. Squaring we get, 
$$Q^2 = |q_1\rangle \langle q_1| + |q_2\rangle \langle q_2| = I$$
by the completeness relation. Plugging this into the equation above, we get:
$$(2I + RQ+QR) \otimes I + (2I -RQ-QR) \otimes I + (I-I)\otimes ST + (I-I)\otimes TS = 4I$$
$$(4I) \otimes I + (RQ+QR-RQ-QR)\otimes I = 4I$$
$$4I = 4I$$
And we are done. (Tensor product properties are on page 73 of Nielsen & Chuang 10th anniversary edition)
A: Not a full answer, but using the identities which I mention in the comments, we can simplify the right side as follows:
$$
4I + [Q,R] \otimes [S,T] = \\
4I + (2i(\vec q \times \vec r) \cdot \vec \sigma) \otimes (2i(\vec s \times \vec t) \cdot \vec \sigma ) = \\
4I - 4 ((\vec q \times \vec r) \cdot \vec \sigma) \otimes ((\vec s \times \vec t) \cdot \vec \sigma )
$$
I would imagine that a similar (albeit longer) simplification can be done on the left side.

Same simplification, applied to the left:
$$
\begin{align*}
(Q\otimes S + R \otimes S &+ R \otimes T - Q \otimes T)^2 = 
\\ &\qquad Q^2 \otimes S^2 + R^2 \otimes S^2 + R^2 \otimes T^2 + Q^2 \otimes T^2\\
&\quad + QR \otimes S^2 + QR \otimes ST - Q^2 \otimes ST\\
&\quad + R^2 \otimes ST - RQ \otimes ST - RQ \otimes T^2\\
& \quad + \text{[previous two lines, with all matrix products reversed]}\\
&= I \otimes I + I \otimes I + I \otimes I + I \otimes I\\
& \quad + [(q \cdot r)I + (q \times r)\cdot \sigma] \otimes I\\
& \quad + [(q \cdot r)I + (q \times r)\cdot \sigma] \otimes [(s \cdot t)I + (s \times t)\cdot \sigma]\\
& \quad - I \otimes  [(s \cdot t)I + (s \times t)\cdot \sigma]\\
& \quad + I \otimes [(s \cdot t)I + (s \times t)\cdot \sigma]\\
& \quad - [(r \cdot q)I + (r \times q) \cdot \sigma] \otimes [(s \cdot t)I + (s \times t) \cdot \sigma]\\
& \quad - [(r \cdot q)I + (r \times q) \cdot \sigma] \otimes I\\
& \quad + \text{[previous six lines, with all vector products reversed]}
\end{align*}
$$
From there, you should be able to cancel a lot to find that we eventually end up with something matching the simplified right side above.
