= 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 $q'(x)=-1/x^2+2$ and $q''(x)=2/x^3>0$. Thus the unique stationary point is the unique global <minimum>, at $x=1/\sqrt2$. Restoring $y$ gives
$$
\boxed{\operatorname*{arg\,min}=\{(1/\sqrt2,\sqrt2)\},\qquad\min=2\sqrt2.}
$$
As a direct global check, the <arithmetic-geometric mean inequality> gives $1/x+2x\ge2\sqrt2$, with equality exactly at the same point.
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