Ytukyg

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+6lambda_1+lambda_2(1-f)=0 $ and $2f+lambda_1-3lambda_2=0$

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.

Surely:$$2x=lambda_1+lambda_2yz=lambda_1-{lambda_2over x}$$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$$u_1={lambda_1+ sqrt{lambda_1^2-4lambda_2}over 4}\u={lambda_1- sqrt{lambda_1^2-4lambda_2}over 4}$$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 $x=yne z$
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 $lambda_2=-2$).