# An elementary inequality for three complex numbers

The following problem arose in asymptotic analysis of difference equations.

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?

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 $$1$$.

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.

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.

• A virtuoso performance! Oct 27, 2020 at 13:15