pulp program for the the following constraint min(a,b) > min(x,y)

Ok. So, the first answer I posted (deleted) was a bit hasty and the logic was faulty for the relationship described. This is (hopefully) correct! ;)

max() and min() are nonlinear, so we need to linearize them somehow (with helper variable) and some logic to relate the 2 minima, which (below) can use a binary helper variable and a Big-M constraint.

in pseudocode:

a, b, x, y : real-valued variables
ab_min : real-valued variable
x_lt_y : binary variable, 1 implies x <= y, 0 else

M = some suitably large constant, depending on the max range of a, b, x, y

new constraints:

ab_min <= a
ab_min <= b
ab_min >= x - (1 - x_lt_y) * M
ab_min >= y - (x_lt_y) * M

Logic:

• We find the minimum of a, b with ab_min.
• We need "upward pressure" from the min(x, y)... So we know that ab_min must be greater than either x or y, or possibly both. For the "or" constraint, we use the binary logic above and multiply it by the "large constant" to make the other constraint trivial.

I would reformulate this as:

z <= x
z <= y
z >= x - Mδ
z >= y - M(1-δ)
a >= z + 0.00001
b >= z + 0.00001
δ ∈ {0,1}

One extra binary variable δ is needed, and one continuous variable z. M is a large enough constant to be chosen with care.