Reduce and FullSimplify don't reduce fully -- how to check if candidate solutions hold

Question

I have a big system of equations and want to check if some candidate solutions hold.

Ideally, I want to have an expression that returns 'True' if my candidate solution hold and 'False' otherwise.

Based on previous posts, I decided to use Reduce[my_equtions && candidate_solutions && assumptions_on_variables]

However, Mathematica doesn't fully reduce the system to True / False -- even after using FullSimplify, some trivial terms remain in the output. Especially when my_equations are multiple, long equation, it is hard to find spot manually if the output should be true or not.

Consider as a MWE the system $x a^2 + y b^2 + (-x - y) ab == 0$ with the assumption $y>x>1$ over the reals. We have candidate solutions $a=1/x$, $b=1/y$ , which clearly solve the system.

However,

Reduce[x a^2 + y b^2 + (-x - y) ab == 0 &&  a == 1/x && b == 1/y && 
   y > x > 1, {a, b}, Reals] // FullSimplify

Outputs:

1 < x < y && ab == 1/(x y) && a == 1/x && b == 1/y

Ideally, I would get True or candidate_solutions && assumptions_on_variables. In any case, the part

ab == 1/(x y)

should not be in there, as these terms become very annoying when the system of equations is large.

My questions:

1. How can I make Reduce to output simply True or candidate_solutions && assumptions_on_variables

2. Is there a better command to test if given candidate solutions fulfill a system of equations under certain assumptions?

Answer 1

Simplify[
 x a^2 + y b^2 + (-x - y) a*b == 0, {a == 1/x, b == 1/y, y > x > 1}]

True.

Answer 2

Provide the conditions to FullSimplify too.

eq = x a^2 + y b^2 + (-x - y) a b == 0;

cond = a == 1/x && b == 1/y && y > x > 1;

Reduce[eq && cond, {a, b}, Reals] // FullSimplify[#, cond] &

(*   True   *)