 Solution

In local holomorphic coordinates, write the two positive singular values of $Df$ as $s_1\ge s_2$. Preservation of orientation gives $J_f=s_1{s}_2$, and quasiconformality gives $s_1/s_2\le K$.

For an admissible conformal metric $\rho |dw|$ on $Y$, define a density on $X$ by $\sigma (z)=\rho (f(z))s_1(z)$. Along each path,so $L_{\rho }(f(\Gamma ))\le L_{\sigma }(\Gamma )$. The change of variables formula and $s_1^2\le K{J}_f$ giveAlso $s_1^2\ge J_f$, so $A_{\sigma }\ge A_{\rho }>0$. Thus $\sigma$ has positive finite area and is admissible, andTaking the supremum over $\rho$ provesApplying the same argument to $f^{-1}$, which has the same bound on its maximal dilatation, also gives $K^{-1}\lambda (\Gamma ,X)\le \lambda (f(\Gamma ),Y)$. The inequalities remain valid for extended extremal lengths; no extremizing density need exist.