= Solution
The <Euler-Lagrange equations> give
$$
\boxed{\begin{aligned}
(m_1+m_2+m_3)\ddot L_1+m_2\ddot L_2+m_3\ddot L_3+k(L_1-L_0)&=(m_1+m_2+m_3)g,\\
m_2(\ddot L_1+\ddot L_2)+k(L_2-L_0)&=m_2g,\\
m_3(\ddot L_1+\ddot L_3)+k(L_3-L_0)&=m_3g.
\end{aligned}}
$$
For example, differentiating $T$ with respect to $\dot L_1$ collects contributions from all three climbers, because changing $L_1$ moves all of them. At <equilibrium>, all accelerations vanish, so
$$
\boxed{\widehat L_1=L_0+\frac{(m_1+m_2+m_3)g}{k},\qquad\widehat L_2=L_0+\frac{m_2g}{k},\qquad\widehat L_3=L_0+\frac{m_3g}{k}.}
$$
The top rope supports the total weight, while each lower rope supports only its own climber. These extensions are positive, consistent with the taut-rope tension law.
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