$min\{x^2+y^2+z^2 : (x,y,z) \in\mathbb{R}^3\}$ with $z\neq 0, xy=\frac{1}{z}$
I have to use the Lagrange multipler here.
This is what I've tried so far:
Let $g(x,y,z)= xyz- 1$
$∇f=λ∇g$ is $(2x, 2y, 2z)=λ(yz, xz, xy)$
Since $xyz=1$, so $yz=\frac{1}{x}$. Then $λ=2x^2=2y^2=2z^2$.
$2x^2-2y^2=0$, so $(x+y)(x-y)=0$
$2y^2-2z^2=0$, so $(y+z)(y-z)=0$
$2z^2-2x^2=0$, so $(z+x)(z-x)=0$
$x=±y$, $y=±z$, $z=±x$
Since $xyz=1$, the minimum would be $(1, -1, -1), (-1, 1, -1), (-1, -1, 1)$
Is this correct? Also the question states 'First show that you can use the Lagrange multiplier here'. Is there a specific condition for using the Lagrange multipier?
Yes, your method is correct.
And there are, in fact, some conditions under which the method of Langrange multipliers can be used. I don't know the clear reason why they are imposed as such, I have been told the proof of validity of this method involves advanced calculus, and I am only a first year undergrad student.
First of all, the number of constraints (that is, the number of the equations available) should be less than the number of variables in the given scalar equation which is to be maximized or minimized.
[A loose logic to this maybe that, if there are more constraints, the point will become predefined as the set of equations would become solvable. But this is extremely loose and has a lot of loopholes.]
In your question, you have only 1 constraint, $g(x,y,z) = xyz-1 = 0$; while you have 3 variables $x, y, z$ in your target function $f$. So this condition is satisfied.
Next, the determinant of Jacobian matrices of the constraint equations should not all be zero at the extreme point.
Here, you have only one constraint equation, whose Jacobian matrix has non-zero determinant value at the obtained point.
Thus, it is correct to use the method here.
Hope it helped