Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics and probability resit / 4 / e / Solution

Both terms $x^6$ and $\frac{3}{2}{y}^4$ are nonnegative, and they vanish simultaneously only at $(0,0)$. That point satisfies $x^4+2y^2\le 3$. Consequentlyand the global minimum is unique. No Lagrange multiplier calculation is needed: nonnegativity gives a lower bound attained at a feasible point.