My solution is rough and takes a few leaps, but I think it's sufficent.
Start with the middle terms, c and d, and find the conditions necessary to build a and b from them.
Let's say c is the geometric mean and d the arithmetic. Then ab = c^2 and a+b/2 = d. Put another way, there is some i such that a = d+i and b = d-i, and since ab = c^2,
(d+i)(d-i) = c^2,
d^2 - i^2 = c^2,
c^2 + i^2 = d^2.
This is equivalent to saying that c is a leg and d is the hypotenuse of some pythagorean triple. We can test this out with the {3,4,5} triangle.
d=5, c=4, (i=3) leads to a=8, b=2. Also,
d=5, c=3, (i=4) leads to a=9, b=1. Each give the desired a/g means.
The question then becomes, can c and d produce e and f? Or, does there exist a pythagorean triple for which the hypotenuse and either leg have integral geometric and arithmetic means?
For this, we look at the pythagorean triple generator:
A = m^2 - n^2, B = 2mn, C = m^2 + n^2.
It is known that in a reduced pythagorean triple, exactly one of A and B is odd, and C is odd. Therefore C must be paired with the odd leg, which is always A, since B is even. Then it will have an integer for an arithmetic mean.
So we want to know if there exists m and n such that m^2 + n^2 and m^2 - n^2 has an integral geometric mean.
(m^2+n^2)*(m^2-n^2) = m^4 - n^4. Can this be a perfect square?
m^4 - n^4 = x^2
x^2 + n^4 = m^4
It has been proved that equations of the form r^2 + s^4 = t^4 have no solutions.
Therefore, there are no integer solutions to this problem.