The following problem arose in asymptotic analysis of difference equations.




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Numerical maximization suggests that for all nonzero complex numbers $a,b,c$ we have
$$h\big(r(a,b,c),r(b,c,a),r(c,a,b)\big)\le2,\tag{1}$$
where
$$r(a,b,c):=\Big|\frac{a b + a c - b c}{a^2}\Big|$$
and $h(\cdot,\cdot,\cdot)$ is the harmonic mean,
with the equality in (1) iff $a,b,c$ are the vertices of an equilateral triangle centered at $0$.


Is this true?




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**Remark 1:** If the above conjecture is true, then obviously it will also hold if the harmonic mean $h\big(r(a,b,c),r(b,c,a),r(c,a,b)\big)$ is replaced by $\min\big(r(a,b,c),r(b,c,a),r(c,a,b)\big)$.


**Remark 2:** For nonzero real $a,b,c$, it appears that the best upper bound $2$ on $h\big(r(a,b,c),r(b,c,a),r(c,a,b)\big)$ should be replaced by $3/2$, "attained in the limit" iff two of the three real numbers $a,b,c$ are equal to each other while the remaining one goes to $0$.


**Remark 3:** For nonzero real $a,b,c$, Mathematica says that the best upper bound on $\min\big(r(a,b,c),r(b,c,a),r(c,a,b)\big)$ is $1$.




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A correct proof of the main, "harmonic-complex" conjecture or of the conjecture stated in Remark 2 or a "human" proof of the fact stated in Remark 3 would be enough for acceptance.



An elementary inequality for three complex numbers
I will prove the original inequality.


First, performing the change of variables $x=1/a$, etc., and inverting the harmonic mean, we need
$$
 \sum \left|\frac{yz}{x(y+z-x)}\right|\geq \frac32.
$$
Next, denoting $p=y+z-x$, etc., we transform the inequality to
$$
 \sum\left|\frac{(p+q)(p+r)}{p(q+r)}\right|\geq 3,
$$
or
$$
 \sum\left|2+\frac{(p-q)(p-r)}{p(q+r)}\right|\geq 3.
$$
Now we perform the last change
$$
 u= \frac{(p-q)(p-r)}{p(q+r)}, \quad\text{etc.,}
$$
and notice that
$$
 \sum\frac 1u=\sum \frac{p(q+r)}{(p-q)(p-r)} =-1
$$
(which is just a three-variable identity). Thus, it suffices to show that
$$
 \sum |2+u|\geq 3 \quad \text{whenever}\quad \sum\frac1u=-1.
$$


Denote $|2+u|,|2+v|,|2+w|$ by $r\_{1,2,3}$, respectively.


Assume first that $r\_i\leq 2$ for all $i$. The inverse image of $|2+z|=r$ (for $r\leq 2$) is a circle whose rightmost point is $-1/(r+2)$, hence, say,
$$
 \Re \frac1u\leq - \frac1{r\_1+2}.
 \qquad(\*)
$$
Adding up three such inequalities, we obtain
$$
 -1\leq-\sum\frac1{r\_i+2}
 \leq-\frac 9{\sum(r\_i+2)},
$$
whence $\sum(r\_i+2)\geq 9$, as desired.


Assume now that, say, $r\_1>2$ (but $r\_1<\sum r\_i<3$, arguing indirectly). Then $\frac1u$ lies outside the disk having diameter $[-1/5,1]$ (as $|2+u|<3$). On the other hand, we have $r\_2+r\_3<1$, so $\frac1v, \frac1w$ lie in the open disks having diameters
$$
 \left[-\frac 1{2-r\_i},-\frac1{2+r\_i}\right].
$$
Since
$$
 \frac1{2-r\_2}+\frac1{2-r\_3}\leq \frac32
 \quad\text{and}\quad
 \frac1{2+r\_2}+\frac1{2+r\_3}\geq \frac45,
$$
point $\frac1u =-1-\frac1v-\frac1w$ lies in the open disk having diameter
$$
 \left[-1+\frac45,-1+\frac32\right]
 = \left[-\frac15, \frac12\right].
$$
which contradicts the previous region indicated for $\frac1u $.


Equality arises only for $u=v=w=-3$, which seems to be rolled back easily.


