Minimization/Sum of Squares in Real Variables

Rearranging $4x^2 + 4y^2 - 4xy - 6y + 3 = 0$ as a sum of squares: $(2x - y)^2 + 3y^2 - 6y + 3 = 0$ $(2x - y)^2 + 3(y - 1)^2 = 0$

Since $x, y$ are real numbers, both square terms must independently equal zero:

1. $3(y - 1)^2 = 0 \implies y = 1$
2. $(2x - y)^2 = 0 \implies 2x = y = 1 \implies x = \frac{1}{2}$

Thus, $4x + 5y = 4\left(\frac{1}{2}\right) + 5(1) = 2 + 5 = 7$.