I trying to solve optimization problem
$\min\underset{x \in C} f(x) = \| x - 1\|^2_2$, where $C = \{x \in \mathbb{R}^3 | x_1 + x_2 + x_3 = 1, x_1 > 0, x_2 > 0, x_3 > 0\}$.
What I did:
I found Lagrangian of $f$: $$L(f) = \frac{1}{2}\|x - 1\|^2_2 + \mu(x_1 + x_2 + x_3 - 1) - \lambda_1 x_1 - \lambda_2 x_2 - \lambda_3 x_3 = \frac{1}{2} ((x_1 - 1)^2 + (x_2)^2 + (x_3)^2) + \mu (x_1 + x_2 + x_3 - 1) - \lambda_1 x_1 - \lambda_2 x_2 - \lambda_3 x_3$$ Since $x_1 > 0, x_2 > 0, x_3 > 0$ and given $\lambda_1, \lambda_2, \lambda_3 \neq 0$ then $\inf L(X, \nu, \lambda_1, \lambda_2,\lambda_3) = -\infty $, hence $\lambda_1, \lambda_2, \lambda_3 \leq 0$ $$\Delta_{x_i} L = \frac{1}{2} \dot 2 (x_i - 1) + \mu = 0 \rightarrow x_i = 1 - \nu - \lambda_i $$
$ min L(x, \nu, \lambda_1, \lambda_2, \lambda_3) = \begin{cases} -\frac{3}{2} + 2\mu, where \: \lambda_1, \lambda_2, \lambda_3 = 0\\ -\infty, where \: \lambda_1, \lambda_2, \lambda_3 =\neq 0 \end{cases} $
I suggest that $x_1 = x_2 = x_3 = \frac{1}{3}$, but I don't know that I'll do next...
