The problem is find the minimum value of $x^2+y^2+z^2$ subject to the condition $x+y+z=1$ and $xyz+1=0$.
Let $f(x,y,z)=x^2+y^2+z^2$, then after some calculation I got this two equations:
$4+6\lambda_1+\lambda_2(1-f)=0 $ and $2f+\lambda_1-3\lambda_2=0$
Now I can solve these two equations to find $\lambda_1$ and $\lambda_2$ in terms for $f$. Now I cant understand how to proceed. Please help with explanation.
On
Surely:$$2x=\lambda_1+\lambda_2yz=\lambda_1-{\lambda_2\over x}$$or equivalently$$2x^2=\lambda_1 x-\lambda_2\\2y^2=\lambda_1 y-\lambda_2\\2z^2=\lambda_1 z-\lambda_2$$the equation $2u^2-\lambda_1 u+\lambda_2=0$ has two roots as following$$u_1={\lambda_1+ \sqrt{\lambda_1^2-4\lambda_2}\over 4}\\u={\lambda_1- \sqrt{\lambda_1^2-4\lambda_2}\over 4}$$Hence of symmetry we consider 2 cases:
case 1: $x=y=z$
this case is impossible since $x+y+z=1$ and $xyz=-1$ don't hold simultaneously
case 2 $x=y\ne z$
in this case $$2x+z=1\\x^2z=-1$$The only answer is $x=y=1,z=-1$ with $\lambda_1=0\\\lambda_2=-2$ and because of symmetry, all the answers are:$$(1,1,-1)\\(1,-1,1)\\(-1,1,1)$$
I don't see where those two equations come from. Applying the method of Lagrange multipliers to your problem, you should have obtained a system of $5$ equations and $5$ unknowns:$$\left\{\begin{array}{l}2x=\lambda_1+\lambda_2yz\\2y=\lambda_1+\lambda_2xz\\2z=\lambda_1+\lambda_2xy\\x+y+z=1\\xyz=-1.\end{array}\right.$$There are only $3$ ponts in $\mathbb{R}^3$ which are solution of this system: $(-1,1,1)$, $(1,-1,1)$, and $(1,1,-1)$ (and, in each case, $\lambda_1=0$ and $\lambda_2=-2$).