TITLE: If my average speed is 50 mph, is it better to go 10% faster or 5 mph faster? (These are not the same)
QUESTION [14 upvotes]: BACKGROUND
Suppose we have a journey between $A$ and $B$ where different sections of roads have different speed limits and our average speed is $50$ mph.
If when I repeat the journey I travel $10\%$ faster at each point, then clearly my average speed will be $10\%$ faster and I will arrive in $1/1.1$ of the original journey time.
On the other hand, if I travel $5$ mph faster at each point, then the average speed may be $10\%$ faster, or it may be different.  (For example, if there is a incredibly slow section of road with a limit of $1$ mph, then it takes ages to go along it at $1.1$ mph, but hardly any time at all when I go at $1+5$ mph.)

Question
When the average speed is $50$ mph, does the option of going $5$ mph faster always beat (or equal) the going $10\%$ faster option?  If so, is there a simple proof?  (This is just a question I was pondering while in a car so I don't know the answer or even whether this is a trivial or hard question...)
What I've tried
I've written a simple Python program to simulate this and experiments indicate that it is always better to go $5$ mph faster (unless all speeds are identical in which case it makes no difference.)
To try and prove it I imagined that there were $50$ sections of road, each $1$ mile long with speed limit $v_i$.  As the average speed is $50$ mph, I know that the total time is $1$ hour and so:
$$
 \sum_{i=1}^{50} { 1 \over{v_i} } = 1
$$
I wanted to show:
$$
 \sum_{i=1}^{50} { 1 \over{v_i + 5} } <= {10 \over 11}
$$
On wikipedia I found that the harmonic mean is less than the arithmetic mean which felt useful, but my attempts to use this got nowhere.
Can anyone find a counter example or perhaps point me in the right direction?

REPLY [8 votes]: When distance is fixed and time is variable, it is often more natural to work with the reciprocal of speed, which I'll call "slowness" and denote $w=1/v$. I'll state a few easily verified properties of slowness:

The time taken for a journey from $0$ to $X$ is $\int_0^X w\,\mathrm dx$.
The average speed $\bar v$ over a journey is the reciprocal of the average slowness $\bar w$.
The average slowness is $\bar w=\int_0^X w\,\mathrm d\mu$, where $\mathrm d\mu=\mathrm dx/X$.

We also have $\int_0^X\mathrm d\mu=1$, which will be needed later.
The problem is to compare two possible journeys: (I) where you increase your speed to $v'=kv$, and (II) where you increase your speed to $v''=v+\delta$, with the condition that $k\bar v=\bar v+\delta$.
In (I), your average slowness becomes $\bar w'=\bar w/k$.
In (II), your slowness becomes
$$w'' = g(w) = \frac1{1/w+\delta},$$
which you can verify is a concave function of $w$. Jensen's inequality therefore implies that
$$g\left(\int_0^X w\,\mathrm d\mu\right)\ge\int_0^X g(w)\,\mathrm d\mu.$$
The left-hand side is $g(\bar w)=\bar w'$, the average slowness of (I). The right-hand side is $\int_0^X w''\,\mathrm d\mu=\bar w''$, the average slowness of (II).
Therefore, (I) is at least as slow as (II), with equality holding only if $w$ is constant almost everywhere.