Just got out of classes and the tutor couldn't find out what was wrong (though he solved it his way):
Find the points of the curve that is the intersection of equations
$$g^1: x^2 + y^2 -xy - z^2 = 1$$
$$g^2: x^2 + y^2 = 1$$
closest to the origin.
Well, that's $$\min_{\left(x,y,z\right)\in\mathbb R^3} (x^2+y^2+z^2)\quad\text{st } g^1, g^2$$
Both restrictions have to be satisfied, than the restrictions reduces to $z^2 = -xy$. Well, if the restrictions are true, than the problem reduces to
$$\min_{\left(x, y, z\right)\in\mathbb R^3}( 1 - xy)$$
If $x$ is minimum, then it has to follow that $1-y =0$ and $1-x=0$, so I get $x=1$, $y=1$ and $z^2 = -1$, and that's a problem, cause z has to be a real value (and dont satisfy restrictions).
I understood Lagrange will give me the right results, but it's cumbersome to do with two restrictions. Any clues what I got wrong?
Here's an elementary (no differential calculus) solution: name $(P_\lambda)$ the horizontal plane having $z=\lambda$ as cartesian equation ($\lambda\in\Bbb{R}$). Now consider any $M\in (g_1\cap g_2)\cap(P_\lambda)$. By Pythagoras' theorem the distance from $M$ to the origin is $OM=\sqrt{1+\lambda^2}\geqslant1$.
If you take $\lambda=0$, any $M\in g_1\cap g_2$ with $z=0$ is such that $\mathbf{OM=1}$, which thus achieves a minimum. Now if you report $z=0$ in the equation for $g_1$ and $g_2$, you easily find out that the solutions are given by $xy=0$ and $x^2+y^2=1$, thus $(x,y)=(\pm1,0)$ or $(0,\pm1)$.
Conversely it is immediately checked that the four points $(\pm1,0,0)$ and $(0,\pm1,0)$ belong to $g_1\cap g_2$.