 Solution

Eliminate the constraint by writing $y=1/x$, with $x>0$. The logarithms cancel, so the objective becomes $q(x)=1/x+2x$. Its derivatives are $q^{\prime }(x)=-1/x^2+2$ and $q^{\prime \prime }(x)=2/x^3>0$. Thus the unique stationary point is the unique global minimum, at $x=1/\sqrt{2}$. Restoring $y$ givesAs a direct global check, the arithmetic-geometric mean inequality gives $1/x+2x\ge 2\sqrt{2}$, with equality exactly at the same point.