The question is this: if we have a set of any random consecutive numbers, for example {1, 1.2, 4.2, 4.8, 5.6, 7.4, 9.8} then how can we prove that calculating the ratio between each of the numbers and picking the largest ratio will give the largest percentage error between the average of the two numbers and the numbers themselves.
Additionally we define the domain of the set to be $$x > 0$$
In this example, you would take the ratio 1.2/1 = 1.2, 4.2/1.2 = 3.5, 4.8/4.2= 1.143..., 5.6/4.8=1.161..., 7.4/5.6= 1.321..., 9.8/7.4=1.324... .
We can see that the largest ratio is 3.5, so we assume that the average of 4.2 and 1.2 = 2.7, will have the largest percentage error w.r.t. either 4.2 or 1.2.
The percentage error would be equal to the difference between the average and either of the two numbers (which in any case will be equal to half the difference between the two numbers) divided by the average itself:
Average: $$((4.2+1.2)/2)$$ Percentage error: $$\frac{((4.2-1.2)/2)}{((4.2+1.2)/2)}$$
Now, instead substituting numerical values for letters,the problem boils down to the following:
Prove that the greatest or least value of $$x/y$$ will always give you the greatest or least value of $$\frac{((x-y)/2)}{((x+y)/2)}$$ which is also equal to $$\frac{(x-y)}{(x+y)}$$
Assume the values are positive.
Consider the function $$F(z)=\frac {z-1}{z+1}$$
It is easy to differentiate: $$F'(z)=\frac 2{(z+1)^2}$$ The derivative is strictly positive so the function is strictly increasing (at least for $z>-1$). Thus it reaches its maximum when $z$ is maximal, and its minimum when $z$ is minimal. But taking $z=\frac xy$ we see that this is your "maximal error" function.