Prove/Disprove $|1 + z_1| + |1 + z_2| + |1 + r z_1 z_2| \ge 1 + \min(r, 1/r)$ 
Let $r$ be a positive real number.
Let $z_1, z_2 \in \mathbb{C}$. Prove or disprove that
$$|1 + z_1| + |1 + z_2| + |1 + r z_1 z_2| \ge 1 + \min(r, 1/r).$$

I came up with this problem after I saw this question ($r = 1$):
Prove $\left|1+z_1\right| +\left|1+z_2 \right| + \left|1+z_1z_2\right|\geq 2$
I used Mathematica to do some numerical experiment which supports the claim.
Also, if $0 < r < 1$,
$\mathrm{LHS} = \mathrm{RHS}$ occurs when $z_1 = -1, z_2 = -1$;
if $r \ge 1$,
$\mathrm{LHS} = \mathrm{RHS}$ occurs when $z_1 = -1, z_2 = 1/r$.
 A: I will be using $|a|+|b|\ge|a\pm b|$ many times.
Four cases, depending on $r$ and $z_2$.
If $0<r<1$ and $|z_2|\ge1$ then
$|1 + z_1| + |1 + z_2| + |1 + r z_1 z_2|$
$\ge|1+z_1|+|z_2||1-rz_1|$
$\ge |1+z_1|+|1-rz_1|$
$=|r+rz_1|+|1-rz_1|+(\frac{1}{r}-1)|r+rz_1|$
$\ge |r+1|+(\frac{1}{r}-1)|r+rz_1|$
$\ge |1+r|=1+r=1 + \min(r, \frac{1}{r})$
The other three cases are just variations of the same idea.
If $0<r<1$ and $|z_2|<1$ then
$|1 + z_1| + |1 + z_2| + |1 + r z_1 z_2|$
$\ge|z_2||1 + z_1| + |1 + z_2| + |1 + r z_1 z_2|$
$\ge|z_1 z_2 -1|+|1 + r z_1 z_2|$
$=|rz_1 z_2-r|+|1 + r z_1 z_2|+(\frac{1}{r}-1)|rz_1 z_2-r|$
$\ge |-r-1|+(\frac{1}{r}-1)|rz_1 z_2-r|$
$\ge |1+r|=1+r=1 + \min(r, \frac{1}{r})$
If $r\ge1$ and $|z_2|\ge1$ then
$|1 + z_1| + |1 + z_2| + |1 + r z_1 z_2|$
$\ge|1+z_1|+|z_2||1-rz_1|$
$\ge |1+z_1|+|1-rz_1|$
$= |1+z_1|+|\frac{1}{r}-z_1|+(r-1)|\frac{1}{r}-z_1|$
$\ge |1+\frac{1}{r}|+(r-1)|\frac{1}{r}-z_1|$
$\ge |1+\frac{1}{r}|=1+\frac{1}{r}=1 + \min(r, \frac{1}{r})$
If $r\ge1$ and $|z_2|<1$ then
$|1 + z_1| + |1 + z_2| + |1 + r z_1 z_2|$
$\ge|z_2||1 + z_1| + |1 + z_2| + |1 + r z_1 z_2|$
$\ge|z_1 z_2 -1|+|1 + r z_1 z_2|$
$=|z_1 z_2-1|+|\frac{1}{r}+z_1z_2|+(r-1)|\frac{1}{r}+z_1z_2|$
$\ge |-1-\frac{1}{r}|+(r-1)|\frac{1}{r}+z_1z_2|$
$\ge |-1-\frac{1}{r}|=1+\frac{1}{r}=1 + \min(r, \frac{1}{r})$
I took a lot from Carl Schildkraut's nice solution.
