If my average speed is 50 mph, is it better to go 10% faster or 5 mph faster? (These are not the same) 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?
 A: 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.
A: As asked, the question is ambiguous and admits multiple answers. Here is why:
Assume we start at $x=0$ at $t=0$ and then our position is described by $x(t)$ at all times. We then assume that we end up at $X_T$ at time $T$. As a reminder the definition of speed is $v(t)=\frac{dx}{dt}$. 
Ambiguity 1): When you say average speed is 50, you need to specify if the average is taken with respect to time t or position x. As a reminder, the average $<f>$ of a function $f(y)$ from $y=0$ to $Y$ is such that $<f> Y=$area under the curve on the graph of $f(y)$,
or more formally
$$<f>=\frac1 T \int_0^Yf(y)dy$$
so it makes a difference if you consider the speed as a function of x: $v(x)$ or as a function of $t$: $v(t)$. In the op it looks like you are thinking more about it as $v(x)$, however it much more common in physics to think of velocity as $v(t)$ and this is what I will assume from now on. (But note that making the other assumption changes the answer.)
It is therefore given that
\begin{equation}
<v>=\frac{X_T}{T}=50
\end{equation}
Ambiguity 2): There are two different notions of what "repeating the journey" means:
i)I keep traveling for the same amount of time T and then see at which position I ended up
or
ii)I keep traveling until I reach position $X_T$ and then check how much time it took me.
I will assume that what the op has in mind is option ii).

We can now address the question:
Case 1) The average speed is 10% faster than before:
$$<v'>=1.1<v>$$
which gives
$$\frac{X_T'}{T'}=1.1\frac{X_T}{T}\Rightarrow$$
$$\frac{X_T}{T'}=1.1\frac{X_T}{T}\Rightarrow$$
$$T'=\frac{T}{1.1}$$
where in the second line I used that $X_T'=X_T$.
Case 2) We are going faster by 5mph at all times. Then:
$$v_2(t)=v(t)+5$$
and we need to integrate once with respect to t to find
$$x_2(t)=x_1(t)+5t$$
we are looking for a time $t=T_2$ such that we have arrived at our destination $x_2(T_2)=X_T$. In other words:
$$X_T=x_1(T_2)+5T_2$$
this can only be solved for $T_2$ if we know exactly $x_1(t)$ or equivalently the instantaneous velocity at all times in the original trip. If this is known, then $T_2$ can be calculated by the formula above and compared to $T/1.1$ that we derived before.
To sum up, the answer to "what is faster" is not universal and depends on the specific velocity profile that we had during the first trip. By plugging in different functions for $x_2(t)$ above we can construct examples were case 1) take less time or case 2) takes less time.
A: Here is my attempt at trying to formulate your question.
Let $d_i$ represent the distance travelled in segment $i$ and $s_i$ represent the speed in segment $i$. Then total time taken ($T$) is given by:$$T=\sum_{i=1}^n\frac{d_i}{s_i}\tag{1}$$Now, if we increase the speed in each segment by some factor $f$ ($\gt1$) then the new time taken ($T_f$) will be given by:$$T_f=\sum_{i=1}^n\frac{d_i}{f\times s_i}=\frac{1}{f}\sum_{i=1}^n\frac{d_i}{s_i}\tag{2}$$However, if we increase the speed in each segment by some increment $\delta$ then the new time taken ($T_{\delta}$) will be given by:$$T_{\delta}=\sum_{i=1}^n\frac{d_i}{s_i+\delta}\tag{3}$$In order for $T_f$ to be quicker than or equal to $T_{\delta}$ we would need to satisfy:$$\frac{1}{f}\sum_{i=1}^n\frac{d_i}{s_i}\le\sum_{i=1}^n\frac{d_i}{s_i+\delta}$$$$\therefore f\ge\frac{\sum_{i=1}^n\frac{d_i}{s_i}}{\sum_{i=1}^n\frac{d_i}{s_i+\delta}}$$
