Answer
The proof is below.
Work Step by Step
We know that the work is equal to the sum of all of the works done over each of the processes. Thus, we find:
1. The first step is adiabatic, so we find:
$P_1V_1^{\gamma}=P_2V_2^{\gamma}$
Using substitution for $\frac{V_1}{V_2}$, it follows:
$P_2=\frac{P_1}{r}$
This means:
$W=\frac{\frac{P_1}{r}V_2-P_1V_1}{\gamma-1}$
2. Pressure is constant, so this is isobaric. Thus, we find:
$W = -\frac{P_1}{r}(V_3-V_2)$
3. This step is adiabatic, so we find:
$P_1V_1^{\gamma}=P_2V_2^{\gamma}$
$P_f=\frac{P_1V_3}{V_1r}$
Thus, it follows:
$W=\frac{\frac{P_1V_3}{V_1r}V_1-V_3\frac{P_1}{r}}{\gamma-1}$
4. This step is constant volume, so no work is done. Thus, adding all of the individual works together and plugging in $\alpha=\frac{V_3}{V_2}$, we obtain:
$W = \frac{P_1V_1(r^{1-\gamma}(\alpha-1)\gamma-\alpha^{\gamma}+1)}{(\gamma-1)}$
