My bad, I didn't check that last step correctly.
This may not quite be within the MX2 syllabus, but if you differentiate the expression
twice, as a function of

and then as a function of

, it can be shown that there is exactly one point in the unit square
)
that is a stationary point in both cases.
(The derivative actually only needs to be taken once, as

and

are symmetric in the expression.)
This point can be shown to be a local maximum by analysing the two derivatives.
The function attains the value of

at this point, so

must be its maximum value over the unit square. (It's actually a global maximum, but that isn't necessary to show.)