But what if my two bricks had different C values? How would I proof it then?
Just go through the maths with two different C values, C
1 and C
2.
From the form of the expressions it is clear that T
M is the arithmetic mean of T
H and T
C (weighted by the C values), while for ΔS to be zero, T
M would have to be the weighted geometric mean (e.g. in the given example, if T
M = sqrt(T
HT
C), the quantity in brackets is 1, so its log is zero). But the arithmetic mean is always greater than the geometric mean, so ΔS > 0.