Find the maximum and minimum values of $x^2 + y^2 + z^2$ subject to the equality constraints $x + y + z = 1$ and $x y z + 1 = 0$
My try:
Let $u=x^2+y^2+z^2$ $$x+y+z-1=0$$ $$xyz+1=0$$ $$(xdx+ydy+zdz)+m(dx+dy+dz)+n(yzdx+xzdy+xydz)=0$$ $$x+m+yzn=0$$ $$y+m+xzn=0$$ $$z+m+xyn=0$$
Multiplying by $x ,y$ and $z$ then adding above three equations i get $u+m+n=0.¢ What should i do after that.. please help me.. thanks in advance.
On
Let us assume that $x,y,z$ are the roots of a monic, cubic polynomial in the $t$-variable, $$ q(t) = t^3-t^2+ct+1. $$ This polynomial has three real roots iff its discriminant is non-negative, i.e. iff $$ 4c^3-c^2+18c+23\leq 0.$$ We want maximize/minimize $x^2+y^2+z^2 = (x+y+z)^2-2c = 1-2c$ over the previous constraint, but the polynomial $4c^3-c^2+18c+23$ has a unique real root at $c=-1$. It follows that under the constraints $x+y+z=1$ and $xyz=-1$ the quantity $x^2+y^2+z^2$ can take any value $\geq 3$, which is attained at the cyclic permutations of $(1,1,-1)$.
Hint: With $$z=1-x-y$$ we get $$x^2+y^2+(1-x-y)^2$$ and the equation $$xy-x^2y-xy^2+1=0$$ Now you can eliminate $x$ or $y$, and you will get a problem in one variable only.