Solving a weird QCQP using a Lagrange multiplier

Question

Minimize $f(x,y,z)=\dfrac12 \left(x^2+y^2+z^2\right)$ in the set $C = \left\{ (x,y,z) \in \mathbb{R^{3}} \mid xy+xz+yz=1 \right\}$.

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I'm stuck on this question. I've shown that it is a non-bounded closed set, on which $f$ is a coercive function, so the minimum exists. Then I'm using the Lagrange multiplier and it becomes terrible. This is the system I obtain $$ \left\{\begin{matrix} x+\lambda y+\lambda z=0 \\ \lambda x+ y+\lambda z=0 \\ \lambda x+\lambda y+ z=0\\ xy+xz+yz=1 \end{matrix}\right.$$ Is there a way to easily solve the Lagrangian equation?

Answer 1

From my perspective, the optimal strategy is to avoid $\lambda$ entirely whenever possible (typically, with more than one constraint, one cannot avoid the multipliers). From $$(x,y,z) = \lambda(y+z,x+z,x+y)$$ assuming that $x,y,z\ne 0$, we get $$\frac{y+z}x = \frac{x+z}y = \frac{x+y}z.$$ If, for example, $x=0$, then we get $y+z=0$, but the constraint equation will not allow this. So, in fact, none of $x,y,z$ can equal $0$.

The first equation $\dfrac{y+z}x = \dfrac{x+z}y$ leads us to $y^2-x^2=z(x-y)$, so either $x=y$ or $x+y+z=0$. The second equation, similarly, will lead us to the conclusion that either $y=z$ or $x+y+z=0$. On one hand, we have $x=y=z$, and so the constraint equation yields $\boxed{x=y=z=\pm1{\sqrt3}}$. On the other hand, if $x+y+z=0$, the constraint gives $(x+y)^2=xy$, so $0=x^2+xy+y^2=(x+\frac12y)^2+\frac34y^2$ and $x=y=z=0$, which cannot occur.

P.S. I don't find this problem in the least weird. :)

Answer 2

The solution is kind of elusive but can be handled. Regarding the system

$$ \cases{ x + \lambda(y + z)=0\\ y + \lambda(x + z)=0\\ z + \lambda(x + y)=0 } $$

for a generic $\lambda$, has the trivial solution $x=y=z=0$ but this solution doesn't obeys the restriction. So to have a non trivial solution we need

$$ \det\left( \begin{array}{ccc} 1 & \lambda & \lambda \\ \lambda & 1 & \lambda \\ \lambda & \lambda & 1 \\ \end{array} \right) = (\lambda -1)^2 (2 \lambda +1) = 0 $$

Here the $\lambda$ choice can be helped by knowing that

$$ x(x + \lambda(y + z))+y(y + \lambda(x + z))+z(z + \lambda(x + y))=x^2+y^2+z^2+2\lambda(x y + y z+ z x) = 0 $$

then

$$ \lambda = -\frac 12(x^2+y^2+z^2) $$

so $\lambda < 0$ ergo $\lambda = -\frac 12$. At this point we know the objective value for the stationary point $(x^*,y^*,z^*)$ which is

$$ (x^*)^2+(y^*)^2+(z^*)^2=1 $$

and $(x^*,y^*,z^*)$ can be calculated by solving

$$ \cases{ x-\frac{1}{2} (y+z)=0\\ y-\frac{1}{2} (x+z)=0\\ z-\frac{1}{2} (x+y)=0\\ x y+x z+y z-1=0 }$$

obtaining

$$ x^*=y^*=z^*=\pm\frac{1}{\sqrt{3}} $$