System of equations in Lagrange multiplier problem Continuing from Confounding Lagrange multiplier problem:
I'm having trouble solving the system of equations below arisen from a Lagrange multiplier problem where we are to optimize $f(x,y,z) = 4x^2 + 3y^2 + 5z^2$ over $g(x,y,z) = xy + 2yz + 3xz = 6$.
$$ \begin{cases} 8x = \lambda (y + 3z) \\ 6y = \lambda (x + 2z) \\ 10 z = \lambda (2y + 3x) \\ xy + 2yz + 3xz = 6 \end{cases} $$
One suggestion I have got is to eliminate the terms $xy$, $yz$ and $xz$, however I have been unable to figure out how to do so. Help much appreciated!
 A: This kind of systems can be tipically done with multipliyng by "convenient ones":
$$\begin{array}{rl}
(1)&8x=\lambda (y+3z)\,\,\,\,/\cdot x\\
(2)&6y=\lambda (x+2z)\,\,\,\,/\cdot y\\
(3)&10z=\lambda (2y+3x)\,\,\,\,/\cdot z\\
(4)&xy+2yz+3xz=6\\
\end{array}$$
I've selected these "convenient ones" because they'll lead into terms with $xy$, $xz$ and $yz$ on the RHS of equations $(1)$, $(2)$ and $(3)$.
Then
$$\begin{array}{rl}
(1)\Rightarrow& \frac{8x^2}{\lambda}=xy+3xz\,\,\,(5)\\
(2)\Rightarrow& \frac{6y^2}{\lambda}=xy+2yz\,\,\,(6)\\
(3)\Rightarrow& \frac{10z^2}{\lambda}=2yz+3xz\,\,\,(7)\\
\end{array}$$
Summing $(5)+(6)+(7)$,
$$\begin{array}{rcl}
\frac{8x^2}{\lambda}+\frac{6y^2}{\lambda}+\frac{10z^2}{\lambda}&=&2xy+4yz+6xz\\
\Rightarrow\frac{1}{\lambda}(4x^2+3y^2+5y^2)&=&(xy+2yz+3xz)\,\,\,(8)
\end{array}$$
Replacing $(4)$ in $(8)$,
$$(4x^2+3y^2+5y^2)=6\lambda\,\,\,(9)$$
EDIT:
This yields an expression that we will use later. 
Now, let's replace $(1)$, $(2)$ and $(3)$ in $(4)$, simplifying their LHS for $x$, $y$ and $z$, respectively:
$$\frac{\lambda(y+3z)}{8} \frac{\lambda(x+2z)}{6}+2 \frac{\lambda(x+2z)}{6} \frac{\lambda(2y+3x)}{10} +3\frac{\lambda(y+3z)}{8}\frac{\lambda(2y+3x)}{10}=6$$
After a bit of simplifying (and using $(4)$ a couple of times and $(9)$ once) you'll get a cubic equation for $\lambda$:
$$\lambda^2(\lambda+8)=40$$
The only real root for this equation is $\lambda=2$.
Now you can find $x$, $y$ and $z$.
Good luck!!
A: Adding the equations $8x=\lambda(y+3z)$, $6y=\lambda(x+2z)$, and $10z=\lambda(2y+3x)$ gives
$\lambda(4x+3y+5z)=8x+6y+10z$, so either $\lambda=2$ or $4x+3y+5z=0$.
$\textbf{1)}$ If $\lambda=2$, we have $8x=y+6z$ and $6y=2x+4z$, so $4x-y-3z=0$ and $x-3y+2z=0$.
Then $3(4x-y-3z)-(x-3y+2z)=0\implies 11x-11z=0\implies z=x$, and then 
$x-3y+2z=0\implies y=x$.
Substituting into the constraint gives $6x^2=6$, so $x^2=1$ and $x=\pm1$.
Thus points where extrema can occur are $(1,1,1)$ and $(-1,-1,-1)$. 
$\textbf{2)}$ If $4x+3y+5z=0$, then $z=-\frac{1}{5}(4x+3y)\implies xy-\frac{2}{5}y(4x+3y)-\frac{3}{5}x(4x+3y)=6$
$\implies
 5xy-8xy-6y^2-12x^2-9xy=30\implies -12x^2-12xy-6y^2=30\implies$ $2x^2+2xy+y^2=-5\implies(y+x)^2+x^2=-5,$ so there is no solution in this case.
