I have the following problem.
$f(x_1, x_2) := (\xi_1 - x_1)^2 + (\xi_2 - x_2)^2 $
Find $(x_1, x_2)$ that minimizes $f$ with constraints : $x_1 \le x_2, \xi_1 > \xi_2$
My try
I have found
$$-2(\xi_1 - x_1) = \frac{\partial f}{\partial x_1} = < \frac{\partial f}{\partial x_2} = -2(\xi_2 - x_2)$$
but I cannot proceed.
The unconstrained minimiser is clearly $(x_1,x_2) = (\xi_1,\xi_2)$ with minimum $0$. In fact, the function $f$ is the square of the function that describes a circular cone with axis parallel to the third coordinate axis and vertex at $(x_1,x_2,f(x_1,x_2))= (\xi_1,\xi_2,0)$. The level sets are circles, and the minimiser is on the circle that lies tangent to the boundary of the constraint set $x_1 \le x_2$. Thus, this point is the projection of $(\xi_1,\xi_2)$ to the $x_1=x_2$ line. Since $\xi_1 > \xi_2$ we need to subtract some vector in the perpendicular space $\{(x_1,x_2)\in\mathbb R^2:x_1=x_2\}^\perp $. This space is one dimensional, so all its elements are of the form $(c,-c)$. That is to say, for some $c\in\mathbb R$, $$\operatorname*{arg\,min}_{(x,y):x\le y} f(x,y) = (\xi_1 - c,\xi_2+c)$$ where $\xi_1-c = \xi_2 +c$, since $x_1=x_2$. Rearranging, $c = \frac{\xi_1 - \xi_2}2$, so that $$\operatorname*{arg\,min}_{(x,y):x\le y} f(x,y) =\left(\frac{\xi_1+\xi_2}2,\frac{\xi_1+\xi_2}2\right) $$
If you prefer a less 'visual' method, you might want to look into Lagrange multipliers.