Maximum or minimum values of $x^2+y^2+z^2$ subject to the condition $x+y+z=1$ and $xyz+1=0$ Using the method of Lagrange Multipliers I get the equations as:
$1) 2x=\lambda (1)+\mu (yz)$
$2) 2y=\lambda (1)+\mu (xz)$
$ 3) 2z=\lambda (1) + \mu (xy)$
Multiplying $(1)$ by $x$, $(2)$ by $y$ and , $(3)$ by $z$ I get:
$2(x^2+y^2+z^2)=\lambda(1)+\mu(3xyz)$
[Using $3xyz=-3$]
$2u=\lambda-3\mu$ where $u$ is the function to be maximized/minimized. $\lambda$ and $\mu$
are constants.
After this I'm not sure how to proceed to find maximum/minimum value of $u$.
Any suggestions?
 A: $2x - \mu yz = \lambda = 2y - \mu xz\\
2(x-y) = \mu z (y-x)\\
\mu z = -2, \text{ or } x= y$
And we can do similar algebra to show that
$\mu x (y-z) = 2(z-y)$ and $\mu y (z-x) = 2(x-z)$
Either $x = y$ or $y = z$ or $x = z$
If we assume that all are different we would come to the conclusion that $\mu x = \mu y = \mu z = -2$ creating a contradiction.
if $x = y$ then $z = \frac{-1}{x^2}$
$2x -\frac {1}{x^2} = 1\\
2x^3-x^2 - 1 = 0$
which only has one real solution:
$(1,1,-1), (1,-1, 1), (-1,1,1)$ is our solution set.
and $x^2 + y^2 + z^2 = 3$
There is no maximum.
A: There is no maximum. First note that clearly $y^2+x^2+z^2\geq y^2$. So can we always raise $y$ without violating the constraints?
Well substituting the second constraint into the first (by setting $x=-1/yz$ ) and rearranging gives $z^2y-zy+y^2z-1=0$. This is a quadratic equation in $z$ and is always solved by $z=\frac{1}{2}(1-y)\pm\sqrt{\frac{1}{y}+(\frac{1}{2}(1-y))^2}$
Thus we can set $y$ to be as high as we like and then find an $x$ and $z$  that satisfy our constraints and so there is no maximum.
