Series Law for Extremal Length

Proposition
Let $$X$$ be a Riemann surface. Let $$\Gamma_1$$, $$\Gamma_2$$ and $$\Gamma$$ be families of rectifiable curves (or, more generally, families of disjoint unions of rectifiable curves) on $$X$$.

Assume that every $$\gamma\in\Gamma$$ contains a $$\gamma_1\in\Gamma_1$$ and a $$\gamma_2\in\Gamma_2$$ such that $$\gamma_1\cap \gamma_2=\emptyset$$.

Then the extremal lengths of $$\Gamma_1$$, $$\Gamma_2$$ and $$\Gamma$$ satisfy
 * $$ \lambda(\Gamma) \geq \lambda(\Gamma_1) + \lambda(\Gamma_2).$$

Remark
The series law and the parallel law are also referred to collectively as the composition laws of extremal length.

Proof
Let $$\rho_1=\rho_1(z)|dz|$$ and $$\rho_2=\rho_2(z)|dz|$$ be conformal metrics as in the definition of extremal length. We may assume that these are normalized so that $$ A(\rho_j) =L( \Gamma_j, \rho_j )$$ for $$ j \in \{ 1 , 2\}$$.

We define another metric $$\rho=\rho(z)|dz|$$ by
 * $$ \rho(z) := \max( \rho_1(z), \rho_2(z) ).$$

Note that this is a well-defined metric.

By definition, the area form $$\rho^2(z)|dz|$$ satisfies
 * $$\rho^2(z)|dz|^2 = \max(\rho_1(z)^2,\rho_2(z)^2)|dz|^2 \leq (\rho_1(z)^2 + \rho_2(z)^2)|dz|^2.$$

Hence
 * $$A(\rho) \leq A(\rho_1) + A(\rho_2) = L(\Gamma_1,\rho_1) + L(\Gamma_2,\rho_2).$$

On the other hand, let $$\gamma\in \Gamma$$, and let $$\gamma_1$$, $$\gamma_2$$ be as in the assumption. Then

$$ $$ $$

Thus
 * $$ L(\Gamma, \rho) \geq L( \Gamma_1 , \rho_1 ) + L( \Gamma_2 , \rho_2 ). $$

Combining this with the inequality for the area, we see that
 * $$ \frac{L(\Gamma, \rho)^2}{A(\rho)} \geq \frac{(L(\Gamma_1 , \rho_1) + L(\Gamma_2,\rho_2))^2}{L(\Gamma_1,\rho_1) + L(\Gamma_2,\rho_2)} = L(\Gamma_1, \rho_1) + L(\Gamma_2,\rho_2).$$

Taking the supremum over all metrics $$\rho_1$$ and $$\rho_2$$ as above, we see that
 * $$ L(\Gamma) \geq L(\Gamma_1) + L(\Gamma_2), $$

as claimed.