51st Ukrainian National Mathematical Olympiad, 4th Round

Denote $t=x+y$, and put $x=t-y$ into the given inequality. Then we get: $$(t-y{)}^2+3(t-y)y+4y^2-\frac{7}{2}\le 0$$ or, equivalently, $$2y^2+ty+t^2-\frac{7}{2}\le 0.$$ The left-hand side of the last inequality can be considered as a quadratic polynomial in $y$. This polynomial has a positive leading coefficient and at least one non-positive value, so its determinant is non-negative: $$D=28-7t^2\ge 0\Rightarrow t^2\le 4.$$ This proves that $t=x+y\le 2$.