Reconciling $Q=PR$ and $Q=vA$ Say we have a closed loop system, where the flow $Q$ is constant through a pipe.
Then, at one segment of the pipe, we make the radius smaller. This causes the resistance $R$ to be greater, so the pressure $P$ must be higher as well. So, at this stretch of pipe, the fluid is under greater pressure. 
Meanwhile, we've made the radius smaller, so that cross-sectional area $A$ is lower. Then, to compensate, $v$ has to be greater, to keep $Q$ equal.
So, in the same stretch of pipe, the velocity $v$ is greater, but so is the pressure. 
Is this at odds with the Bernoulli principle, which says that when pressure and velocity of a fluid are inversely proportional?
I have a very limited understanding of the Bernoulli equation, so it's very possible that I'm misinterpreting it. But my thought was that if a fluid flows faster through a pipe, the collapsing pressure on that pipe would be greater. However, this seems at odds with the fact that the fluid appears to be under more pressure when it flows faster.
 A: 
I have a very limited understanding of the Bernoulli equation, so it's very possible that I'm misinterpreting it. But my thought was that if a fluid flows faster through a pipe, the collapsing pressure on that pipe would be greater. 

The following derivation will show that the pressure gradient along the flow line of the system depends also on the lengths of the pipes, not only on the diameters of the pipes.
Consider the following diagram for a horizontal pipe (no changes in potential energy):

As the fluid flows the total pressure loss $\Delta p$ is:
$$\Delta p = (p_3-p_2)+(p_2-p_1)$$
Using Darcy-Weisbach, the pressure drop $\Delta p_i$ over each segment of pipe, assuming laminar flow (see edit for turbulent flow) is given by:
$$\Delta p_i=\frac{128\eta L_i Q}{\pi D_i^4}$$
$\eta$ is the viscosity and $Q$ the volumetric through-put. So that:
$$(p_3-p_2)+(p_2-p_1)=\frac{128\eta L_1 Q}{\pi D_1^4}+\frac{128\eta L_2 Q}{\pi D_2^4}$$
So that the overall pressure drop can be calculated:
$$p_3-p_1=\frac{128\eta L_1 Q}{\pi D_1^4}+\frac{128\eta L_2 Q}{\pi D_2^4}$$
Assuming the flow is into atmospheric pressure, then $p_3$ is known and by calculation also $p_1$.
$p_2$ can then be calculated by:
$$p_2=p_1+\frac{128\eta L_2 Q}{\pi D_2^4}$$
The pressure gradient will be something like (schematic, not to scale):


Edit: on turbulent flow.
For turbulent flow the expression for $\Delta p_i$ is replaced with:
$$\Delta p_i=\lambda_i\frac{L_i}{D_i}\frac{v_i^2}{2g}$$
Where $v_i$ is the mean flow velocity and $\lambda_i$ the friction coefficient for turbulent flow, which can be calculated from the equations on this page.
The derivation then proceeds analogously to the case of laminar flow.
A: 
Then, at one segment of the pipe, we make the radius smaller. This
causes the resistance R to be greater, so the pressure P must be
higher as well

No. It is just the opposite. The total energy of a given mass of fluid is constant, being the sum of a potential term $PV$ and a kinetic term $\frac{1}{2}mv^2$.
Exactly because the velocity increases in the narrow region, the pressure decreases.
Considering the friction of the fluid with the pipe walls, we can say that the pressure drop for the same length is greater in that region, what is another issue.
