# 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 $$z_1 = -1, z_2 = 1/r$$.

I will be using $$|a|+|b|\ge|a\pm b|$$ many times.

Four cases, depending on $$r$$ and $$z_2$$.

If $$0 and $$|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 $$0 and $$|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.

• Case 2 has an error in the final line. Commented Jul 21, 2022 at 8:07
• @SuzuHirose Thanks, corrected.
– Dan
Commented Jul 21, 2022 at 8:09