The easy proof is the commutative property of multiplication, and my calculus is too rusty to do it for real.
Here's my best attempt:
Given x, an odd number of containers in scoring position,
let y = the optimal stack height (which you have proofed)
y = x/2
In the case of an odd number of containers, y will be a complex fraction (where x = 11, y = 5 1/2). Since we cannot have 1/2 a container, we must round y either up or down, by adding 0.5 or subtracting 0.5.
We can calculate 2 scores based on (1) adding another container to the stack and (2) not adding another container to the stack.
s1 = (y+0.5) (y-0.5)
s2 = (y-0.5) (y+0.5)
Given the commutative property of multiplication (which states that A*B=B*A), we prove that s1 = s2.