Conditional Extremum, need help finding the extreme points in calculation. Find the conditonal extremums of the following $$u=xyz$$ if $$(1) x^2+y^2+z^2=1,x+y+z=0.$$
First i made the Lagrange function $\phi= xyz+ \lambda(x^2+y^2+z^2-1) + \mu (x+y+z) $, now making the derivatives in respect to x, y, z equal to zero i get the system :
$ \phi_x'=yz+2\lambda x+ \mu $
$\phi _y'=xz+2\lambda y+ \mu $
$\phi _z'=xy+2\lambda z+ \mu$
Using these systems and (1) , 6 points are found. I can't seem to calculate these points. Can anyone help with the calculation, i dont know how to come about the result..
 A: This is easier to achieve along the following lines: assume that $x,y,z$ are three real roots of a polynomial
$$ p(t) = t^3 + at^2 +bt +c.$$
By Vieta's formulas we have $a=0$ and 
$$2b=2xy+2xz+2yz=(x+y+z)^2-(x^2+y^2+z^2)=-1,$$
so 
$$ p(t) = t^3-\frac{t}{2}-u $$
must have three real roots. Since $f(\xi)=\xi^3-\frac{\xi}{2}$ has the following graphics:

in order that $f(\xi)=u$ has three real solutions $u$ must be between the values of $f$ in its stationary points, i.e.:
$$ u\in\left[-\frac{1}{3\sqrt{6}},\frac{1}{3\sqrt{6}}\right].$$
A: $x+y+z=0$ means $x,y,z$ have different signs,but for three ones ,there are two ones have same sign. WLOG, let $xy>0 \implies  xy \le \dfrac{(x+y)^2}{4}$ 
we have $x^2+y^2+(x+y)^2=1 \ge \dfrac{(x+y)^2}{2}+(x+y)^2 \iff (x+y)^2 \le \dfrac{2}{3}$
when $x=y=\pm \dfrac{\sqrt{6}}{6} ,(x+y)^2$ get max and $xy$ get max also.
$u=-xy(x+y)$, 
when $x>0,y>0 \implies x+y>0 \implies xy(x+y)\le \dfrac{(x+y)^3}{4}\le \dfrac{\sqrt{6}}{18} \implies u \ge - \dfrac{\sqrt{6}}{18} $
so the min points are $(\dfrac{\sqrt{6}}{6},\dfrac{\sqrt{6}}{6},-\dfrac{\sqrt{6}}{3})$
when $x<0,y<0 \implies x+y<0 \implies xy(x+y)\ge \dfrac{(x+y)^3}{4}\ge- \dfrac{\sqrt{6}}{18} \implies u \le \dfrac{\sqrt{6}}{18}  $ or premium 
so the max points are $(-\dfrac{\sqrt{6}}{6},-\dfrac{\sqrt{6}}{6},\dfrac{\sqrt{6}}{3})$
