I am not sure how you will apply it to 4 numbers but with 2 numbers here's how you might do it-

a+b = n

m(Arithmetic Mean) = (a+b)/2

let d = |a-m| = |b-m|

Without the loss of generality assume a>=b.

So, a = m+d

and b = m-d

Our Goal- Maximise a*b

To Maximise a*b, we need to Maximise (m+d)(m-d)

(m+d)(m-d) = m^2 - d^2 = (n/2)^2 - (d)^2 = n^2/4 - (d)^2

n^2/4 is a constant, we cannot change it by changing the values of a and b.

So we need to Minimise (d)^2. (d)^2 is 0 for d = 0 which is its lowest possible value.

Therefore d = 0.

But a = m + d. So d = a-m.

a - m = 0 implies that a = m.

(a + b)/2 = m

(a+b)/2 = a

a+b = 2a

b = a.