Intuitively, why do attempts to delay hitting a black hole singularity cause you to reach it faster? In general relativity, proper time is maximized along geodesics. Inside of a black hole, all future-oriented timelike trajectories end at the singularity. Putting these two facts together, we find that any deviation from geodesic free fall decreases the proper time before one hits the singularity, so as Carroll says, "you may as well sit back and enjoy the ride."
[Edit: As Dale points out, the Schwarzchild singularity does not consist of a single spacetime event, so this argument fails in general: one can in fact extend the proper time experienced by a free faller between the event horizon and the singularity to some extent by firing rockets inward. But this cannot occur appreciably in the limiting case where the free fall begins at rest just outside the horizon, which I'll assume to be the case.]
This is of course very much counter to nonrelativistic intuition. In Newtonian gravitation, if you fire your jetback inward, you slow your inward fall and buy yourself more time. Is there any physical intuition for why this isn't the case inside of a black hole (if you start free falling from rest at the horizon)?
 A: My (very limited) intuition for this is that once you cross the event horizon, the singularity is not so much a distant point in space as it is a moment in future time.  
In other words, within the event horizon you're firing your rockets not to avoid some point $(x,y,z)$, but rather to avoid next Thursday.  From here, I use my intuition about time dilation and the fact that geodesics are trajectories of maximum proper time.
I'm by no means a GR expert so if this picture is wrong, corrections are more than welcome :)
A: Here's a partial answer, although it's still pretty formal. First define
$$E := -\left( \frac{2GM}{r} - 1 \right) \frac{dt}{d\tau}, \qquad L := r^2 \frac{d\phi}{d\tau}$$
in the usual Schwarzschild coordinates. If you're free falling, then $E$ and $L$ are constant over your trajectory, but if you can fire your engines then they can change. We can expand out the normalization condition $U \cdot U = -1$ to
$$\frac{1}{2} \left( \frac{dr}{d\tau} \right)^2 - \left(\frac{2GM}{r} - 1 \right) \left(1 + \frac{L^2}{r^2} \right) = \frac{1}{2} E^2,$$
which looks somewhat like the statement of conservation of energy (per unit mass) for a nonrelativistic particle with angular momentum $L$.
But there are two weird aspects to this equation:


*

*When you expand out the product of the two binomials on the LHS, you get a weird term $-2GM L^2/r^3$ that does not appear in the nonrelativistic case. Unlike the usual centrifugal angular momentum barrier, this is a centripetal angular momentum fictitious "force" that actually sucks the particle inward at small radii. This means that angular momentum is actually your enemy, not your friend, for avoiding the singularity - so you don't want to accelerate in a way that increases its magnitude.

*The effective total energy $\mathcal{E}$ isn't the physical mechanical energy $E$, but instead $\frac{1}{2} E^2$. In the standard nonrelativistic case, firing your engines to slow your infall decreases your total mechanical energy and helps you delay getting close to the center. (This may seem counterintuitive at first, because we associate highly negative energies with tightly bound orbits and positive energies with unbound orbits, so you might think you would want to increase your energy. But for the purpose of delaying getting close to the center, you actually want to brake and make your energy more negative, at the expense of trapping yourself deeper in the gravity well overall and spending more time near the center once you finally do get there.) But in the Schwarzschild case, $\mathcal{E} = \frac{1}{2} E^2$ means that your effective energy actually depends non-monotonically on your physical energy: if your physical energy $E$ is negative, then making it even more negative actually increases your effective energy $\mathcal{E}$. This means that minimizing your effective energy requires keeping your physical energy at $E = 0$, which indeed corresponds to the optimal geodesic which begins at rest infinitesimally outside the horizon. Any attempt to brake further will overshoot $E = 0$ and send $E$ negative, which will actually increase your effective energy and hurt you.
A: Actually, it turns out to be incorrect that the optimal strategy is to free fall. There is an optimal strategy for firing your rocket engine which maximizes your proper time from the event horizon to the singularity, and extends it beyond the proper time of a free falling observer. 
Here is a paper that discusses the issue and describes strategies for maximizing the proper time to the singularity:
https://arxiv.org/abs/0705.1029v2
Edit: a TL;DR summary of the paper. An infalling rocket can maximize the proper time to the singularity by first making a burn to match the trajectory of a free falling object which started at rest at the horizon. Once the rocket has matched that specific trajectory, then it should turn off engines. 
A: In a metric, such as
$$ d\tau^2=g_{11}dt^2-g_{22}dr^2 $$
the longest interval $d\tau$ between two events obviously is when $dr=0$ simply due to the sign. This is the rest frame with no motion in space and consequently no time dilation due to motion. Any $dr\ne 0$ would result in a motion with a stronger time dilation and therefore decrease the interval or the proper time.
The radial geometrized Schwarzschild metric inside the event horizon is
$$ d\tau^2 = \left(\frac{r_s}{r}-1\right)^{-1} \,dr^2 - \left(\frac{r_s}{r}-1 \right)\,dt^2\tag{1} $$
Where $r$ is the coordinate time and $t$ is a spatial coordinate orthogonal to time and therefore not pointing to the center. As mentioned above, the longest proper time is when $dt=0 $ and therefore
$$ d\tau^2 = \left(\frac{r_s}{r}-1\right)^{-1} \,dr^2 $$
Or
$$ d\tau =\dfrac{dr}{\sqrt{\dfrac{r_s}{r}-1}} $$
Solving
$$ \tau=-r\sqrt{\dfrac{r_s}{r}-1}-r_s\arctan\left(\sqrt{\dfrac{r_s}{r}-1}\right)+C $$
From $\,r=r_s\,$ to $\,r=0\,$ the longest possible lifetime inside the black hole is
$$ \tau=\dfrac{\pi}{2}r_s=\pi M $$
More rigorously, the bound solution of the geodesic equations for the radial metric $(1)$ yields the following geodesics (where $R$ is the radius, from which the fall starts at rest)
$$ \tau=\dfrac{R}{2}\sqrt{\dfrac{R}{2M}}\left(\arccos\left(\dfrac{2r}{R}-1\right)+\sin\left(\arccos\left(\dfrac{2r}{R}-1\right)\right)\right) $$
And
$$ t=\sqrt{\dfrac{R}{2M}-1}\cdot\left(\left(\dfrac{R}{2}+2M\right)\cdot\arccos\left(\dfrac{2r}{R}-1\right)+\dfrac{R}{2}\sin\left(\arccos\left(\dfrac{2r}{R}-1\right)\right)\right)+ $$
$$ +\, 2M\ln\left(\left|\dfrac{\sqrt{\dfrac{R}{2M}-1}+\tan\left(\dfrac{1}{2}\arccos\left(\dfrac{2r}{R}-1\right)\right)}{\sqrt{\dfrac{R}{2M}-1}-\tan\left(\dfrac{1}{2}\arccos\left(\dfrac{2r}{R}-1\right)\right)}\right|\right) $$
Plotting these functions for the fall from the horizon $r=2M$ confirms no spatial movement $t=0$ (blue line), as well as the maximum proper time $\tau=\pi M$ (green line). Please note that time $r$ on the chart moves from right to left.

In comparison, the next plot represents a fall from $r=5M$ showing time $t$ above the horizon diverging to infinity and showing a fast spatial movement along $t$ inside the horizon causing a stronger time dilation that results in a (roughly twice) smaller value of the proper time $\tau$ between the horizon at $r=2M$ and the singularity at $r=0$.

The charts show that gravity inside a black hole causes a deceleration of moving bodies $\dfrac{d^2t}{dr^2}\lt 0$ and does not accelerate bodies at rest with the speed of $\dfrac{dt}{dr}=0$.
Using these results, we can now visualize the geometry of a Schwarzschild black hole in a spacetime reduced by one dimension

In this diagram, the coordinate $t$ is vertical. Outside the event horizon $t$ represents time; inside the event horizon $t$ represents a spatial dimension that does not point to the singularity. The radial coordinate $r$ is spatial outside the horizon, but represents time inside. Thus the singularity is a line along the spatial dimension of $t$ at the time of $r=0$.
A body falling from the event horizon $A$ has no momentum along the spatial dimension of $t$. Therefore this body is stationary inside and moves only in time along $r$ from $A$ to $B$. Due to the symmetry considerations, this body cannot gain a momentum along the spatial direction of $t$ during the fall. For this reason, a body falling from the event horizon would have the longest possible lifetime inside the horizon, as discussed above. While we call this movement "a free fall", in fact the body remains stationary in space.
A different body in a free fall from infinity or from any point outside would move outside the horizon along the geodesic from $C$ to $D$. Passed the point $D$ time diverges to infinity for an external observer. After crossing the horizon, this body continues moving along the geodesic from $E$ to $F$ (see also the geodesic chart above). Because this body moves in space along the dimension of $t$, the body experiences a time dilation due to motion that shortens its total proper time inside the black hole.
To extend the proper time, the movement along $t$ must be decelerated and stopped, as shown at $G$. After that the body is stationary with no movement in space along $t$ while moving only in time along $r$ from $G$ to $H$. Obviously, provided the deceleration time is negligible, the lifetime of this body is maximized as discussed earlier.
