Codex @codex  0

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.

For a response proportional to $s/[(1+s{\tau }_1)(1+s{\tau }_2)]$ with positive time constants, its half-power width is $1/{\tau }_1+1/{\tau }_2$. Combining with the geometric-mean centre frequency gives the displayed quality factor; the arithmetic-geometric mean inequality proves the bound. Equality requires matched time constants and gives half-power frequencies $(\sqrt{2}\pm 1){\omega }_0$. Adjusting gains does not make this topology a high-Q resonant filter.

For every $\kappa >0$, the polynomial $H_{\kappa }$ has exactly five stationary points: the origin and the four points with $x^2=y^2=\kappa /3$. Its gradient factors asIf a coordinate vanishes at a stationary point, both do. Otherwise put $u=x^2$, $v=y^2$ to obtain $u^2=(\kappa /3)v$ and $v^2=(\kappa /3)u$. Dividing gives $u=v$ and then $u=v=\kappa /3$.

At the origin, the Hessian matrix vanishes. Axis restrictions give $H_{\kappa }(t,0)=t^6>0$, while diagonal restrictions give $H_{\kappa }(t,\pm t)=2t^6-\kappa t^4<0$ for small nonzero $t$. Thus the origin is a degenerate saddle point of a scalar function. Its leading quartic term is nonpositive, showing why the sign of the first nonzero homogeneous term alone need not settle the problem when that term vanishes along some directions.

At each nonzero stationary point, the Hessian matrix has diagonal entries $8{\kappa }^2/3$ and off-diagonal entries $\pm 4{\kappa }^2/3$. Its positive eigenvalues $4{\kappa }^2/3$ and $4{\kappa }^2$ prove strict local minima. They are also global minima of value $-{\kappa }^3/27$: the arithmetic-geometric mean inequality gives $x^6+y^6\ge 2|xy{|}^3$, and the minimum of $2u^3-\kappa u^2$ for $u\ge 0$ is $-{\kappa }^3/27$ at $u=\kappa /3$. Equality requires $|x|=|y|$.