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Map a conformal quadrilateral to a rectangle of width $w$ and height $h$, with the chosen opposite boundary arcs mapped to its vertical sides. Their extremal length is $M=w/h$; the other pair has extremal length $1/M$. The constant conformal metric is extremal. For an arbitrary density, average horizontal joining lengths over height and apply the Cauchy-Schwarz inequality to obtain $L_{\rho }^{2}h\le w{A}_{\rho }$, proving the upper bound. Vertical joining segments prove the dual formula. Conformal invariance of extremal length transports both equalities back to the quadrilateral.

By conformal invariance of extremal length, work first on $1<|w|<R$. For a conformal metric $\rho$, let $L_{\rho }$ be the infimum of lengths of curves joining the two circles, and $A_{\rho }$ its area. Each radial segment is admissible, so the Cauchy-Schwarz inequality gives, for every angle,Integrating in angle gives $2\pi L_{\rho }^2\le A_{\rho }\log R$. Conversely, $\rho (w)=1/|w|$ has shortest joining length $\log R$ and area $2\pi \log R$, attaining the bound. Thus the conformal modulus of an annulus givesThe second formula is the pullback of $c/|w|$ by $\psi$. Pullback preserves curve lengths and metric area, so every displayed positive multiple is an extremal metric in $\Omega$.