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:
$$D = 28 - 7t^2 \geq 0 \quad \Rightarrow \quad t^2 \leq 4.$$
This proves that $t = x + y \le 2$.