How to minimize an equation with absolute values? How would I go about minimizing the expression
$\left(|z_1| + |z_2|\right) \times \left(|z_1 + z_2| + |z_1 - z_2|\right)$
subject to the constraint
$|z_1|^2 + |z_2|^2 = 1$
given that $z_1$ and $z_2$ can be complex numbers?
I thought of trying Lagrange multipliers, but it doesn't seem possible because there are an infinite number of solutions (and solving a 5-equation, 5-variable system is a bit painful).
Any hints on how I could do this?
 A: This is just a quick shot, so I may have made a mistake, but I think the answer is that $z_1=1$ and $z_2=0$ is an optimum, giving a result of $2$.
Since there is symmetry with respect to rotation, fix $z_1$ to be real.
Since there is symmetric treatment of $z_1$ and $z_2$, I suspect that either one of them is zero or else their magnitudes are equal. This isn't certain --- it's also possible that the optimum is two unequal, nonzero magnitudes, and you can arbitrarily choose which one you want to be the larger.
Assuming equal magnitudes, we have $z_1=1/\sqrt{2}$ and $z_2=e^{i\phi}/\sqrt{2}$. The minimum result then occurs when $\phi=0$, giving the claimed result.
Assuming that one has magnitude 0, we also get a result of 2.
To be certain of the result, you really need to take $z_1=\cos\theta$, $z_2=\sin\theta e^{i\phi}$, and minimize simultaneously with respect to $\theta$ and $\phi$.
EDIT: Here's a sketch for a complete proof. The function to be minimized is
$f=\sqrt{2}\sin(\theta+\pi/4)\left(\sqrt{1+\sin2\theta\cos\phi}+\sqrt{1-\sin2\theta\cos\phi}\right)$
where $0 \le \theta \le \pi/2$ and $0 \le \phi \le \pi/2$. (The expression is invalid for $\theta$ in the second quadrant, and points with $\phi$ in the second quadrant give the same result as points in the first quadrant.)
It should be pretty easy to prove that on this rectangle in the $(\theta,\phi)$ plane, $\partial f/\partial\phi \ge 0$, so you only have to search for minima along the edge at $\phi=0$. Along this edge, there are no local minima at differentiable points, but there is a minimum at the nondifferentiable point $\phi=\pi/4$, as well as at the corners. All three of these minima give $f=2$.
A: EDIT: my first try found the maximum rather than the minimum.
Suppose wlog $|z_1| \ge |z_2|$.  Fix $z_1$ and $|z_2|$ for the moment, and write $z_2 = w z_1$.  Then we want to minimize $|z_1 + z_2| + |z_1 - z_2| = |z_1| (|w + 1| + |w - 1|)$ subject to $|w| = |z_2|/|z_1|$.  Geometrically, $|w + 1| + |w - 1|$ is the sum of the distances from $w$ to $1$ and $-1$, which is $2$ on the line segment $[-1,1]$ and greater than $2$ otherwise, and $w$ is on a circle centred at $0$ with radius less than $1$, so the minimum value is attained at $w = \pm |z_2|/|z_1|$.  Now write $|z_1| = \cos \theta$ and $|z_2| = \sin \theta$, $0 \le \theta \le \pi/4$.  We have $|z_1 + z_2| + |z_1 - z_2| = 2 \cos \theta$, so the objective is to minimize $2 (\cos \theta + \sin \theta)\cos \theta$ for $0 \le \theta \le \pi/4$, and it is
easy to see that the minimum value is $0$, attained at both $\theta=0$ and $\theta=\pi/4$.
