Let
i) $\mu = [\mu_1,\mu_2,\mu_3]\in\mathbb{R}^3$, such that $\mu_2 > \mu_1$, $\mu_2 >\mu_3$ fixed,
ii) $\lambda = [\lambda_1,\lambda_2,\lambda_3] \in \mathbb{R}^3$ such that $\lambda_1 \geq \lambda_2 \geq \lambda_3$,
iii) $\alpha\in\Sigma_3$, where $\Sigma_3$ is the simplex of dimension $3$, i.e. $\Sigma_3 = \{\alpha : \alpha_1+\alpha_2+\alpha_3 = 1\}$.
I am stuck in solving the following maximin problem:
$$ \sup_{\alpha} \inf_{\lambda} f(\alpha,\lambda) = \sup_{\alpha} \inf_{\lambda}\sum_{i =1}^{3}\alpha_i(\mu_i -\lambda_i)^2.$$
What I have done: I framed the problem in the constrained optimisation setting and I solved the first part (inf)
$$ \inf_\lambda f(\alpha,\lambda) \\s.t. \lambda_2 -\lambda_1 \leq 0 \\ \;\;\;\;\lambda_3 -\lambda_2 \leq 0$$
formed the Lagrangian $\mathcal{L}(\alpha,\lambda,\rho) = f(\alpha,\lambda) - \rho_1( \lambda_2 -\lambda_1)-\rho_2( \lambda_3 -\lambda_2)$, and gone through the solution of the problem by imposing KKT conditions. I get two different solutions that depends on the values of alpha:
$$ \begin{cases}\lambda_1 = \lambda_2 = \lambda_3 = \bar{\mu} = \sum_{i} \alpha_i\mu_i \text{ if }\mu_1 \geq \bar{\mu} \text{ and } \mu_3 \leq \bar{\mu} \\ \lambda_1 = \lambda_2 = \bar{\mu}_{12} = \frac{1}{\alpha_1 + \alpha_2} \alpha_1\mu_1 + \alpha_2\mu_2, \lambda_3 = \mu_3 \text{ if } \mu_2 \geq \bar{\mu}_{12} \text{ and } \mu_1 \leq \bar{\mu}_{12} \end{cases} $$
When I pass to the second outer optimisation problem (sup) I follow the same approach and treat the two cases independently propagating the KKT constraints in the constraints of the sup problem and adding the equality constraint for the probability simplex on the alphas. However, I cannot find any feasible solution from this problem.
Since I am new to optimisation, and essentially self-taught, I was wondering if I am following the right approach to solve such maximin problem or if there is some easier way to proceed. Any help from more knowledgeable people will be of great help to me since I have been stuck on this problem for a while now.
Thanks to anyone who will try to help in advance.