For any given real numbers such that $\lambda_1\geq\lambda_2\geq\lambda_3\geq\lambda_4\geq\lambda_5\geq\lambda_6$, show that the optimal solution of the problem \begin{align} \mbox{maximize}& -(\lambda_1^+ + \lambda_1^-) + \sqrt{(\lambda_1^+-\lambda_1^-)^2+(\lambda_3^+-\lambda_3^-)^2}\\ \mbox{subject to }& \{\lambda_1^+,\lambda_1^-,\lambda_2^+,\lambda_2^-,\lambda_3^+\lambda_3^-\} = \{\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5,\lambda_6\}\quad\mbox{and}\quad\lambda_i^+\geq \lambda_i^-\quad \mbox{for every}\quad i\in\{1,2,3\} \end{align} is given by \begin{equation} (\lambda_1^+,\lambda_1^-,\lambda_2^+,\lambda_2^-,\lambda_3^+,\lambda_3^-)=(\lambda_4,\lambda_6,\lambda_2,\lambda_3,\lambda_1,\lambda_5) \end{equation}
OBS 1: The proposed solution was obtained numerically by enumerating more than a million instances of sets $\{\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5,\lambda_6\}$ and, for each of them, explicitly evaluating the figure of merit for every possible permutation of the six elements and picking the largest one.
OBS2: Although the result seems to hold for any real values of $\lambda_j$, I am only interested in the case where $\lambda_1\geq\lambda_2\geq\lambda_3\geq\lambda_4\geq\lambda_5\geq\lambda_6\geq 0$ and $\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5+\lambda_6=1$. So, feel free to use these restrictions if that makes the proof any easier.