how prove this nice inequality $\max\{|z_{1}z_{2}|^2-2|z_{1}|^2-|z_{3}|^2,\cdots\}\}\ge 2016$

Question

Let three complex numbers $z_{1},z_{2},z_{3}$,such $$ \{|z_{1}+z_{2}+z_{3}|,|-z_{1}+z_{2}+z_{3}|,|z_{1}-z_{2}+z_{3}|,|z_{1}+z_{2}-z_{3}|\}=\{98,84,42,28\} $$ show that $$\max\{|z_{1}z_{2}|^2-2|z_{1}|^2-|z_{3}|^2,|z_{2}z_{3}|^2-2|z_{2}|^2-|z_{1}|^2,|z_{3}z_{1}|^2-2|z_{3}|^2-|z_{2}|^2\}\}\ge 2016$$(creat by Xi Yong Wang)

Answer 1

Denote complex variables $$ m = z_{1}+z_{2}+z_{3}\\ r_1 = -z_{1}+z_{2}+z_{3} = |r_1| e^{i \phi_1}\\ r_2 = z_{1}-z_{2}+z_{3}= |r_2| e^{i \phi_2}\\ r_3 = z_{1}+z_{2}-z_{3} = |r_3| e^{i \phi_3} $$ Now $m = r_1 + r_2 + r_3$ and (the indices are understood cyclically, i.e. $k=4$ means $k=1$ etc.): $$ |m|^2 = m m^* = (r_1 + r_2 + r_3)(r_1 + r_2 + r_3)^* = \\ = \sum_{k=1}^3 |r_k|^2 + 2 \sum_{k=1}^3 |r_k| |r_{k+1}| \cos(\phi_k - \phi_{k+1}) $$ Now it happens that $$ |m|^2 = \sum_{k=1}^3 |r_k|^2 \quad {\rm{as}} \\ 98^2 = 84^2 + 42^2 + 28^2 $$ Writing $\alpha_1 = \phi_1 - \phi_2$ and $\alpha_2 = \phi_2 - \phi_3$ and $\alpha_3 = \phi_3 - \phi_1 $ hence $$ \alpha_1 + \alpha_2 + \alpha_3 = 0 $$

We have $$ 0= |r_1| |r_2| \cos(\alpha_1)+ |r_2| |r_3| \cos(\alpha_2) + |r_3| |r_1| \cos(\alpha_3) $$

The complex variables $z_k$ are given by $z_k = \frac12 (r_{k+1} + r_{k+2})$, hence $|z_k|^2 = \frac14 (|r_{k+1}|^2 + |r_{k+2}|^2) + \frac12 |r_{k+1}| |r_{k+2}| \cos \alpha_{k+1}$, and with the previous result follows

$$ \sum_k |z_k|^2 = \frac12 \sum_k |r_k|^2 = \frac12 98^2 $$

Now the task at hand is to regard quantities

$$ F_k = |z_{k}|^2 |z_{k+1}|^2-2|z_{k}|^2-|z_{k+2}|^2 \\ = |z_{k}|^2 |z_{k+1}|^2 +2|z_{k+1}|^2 + |z_{k+2}|^2 -2(|z_{k}|^2+ |z_{k+1}|^2+|z_{k+2}|^2) \\ = |z_{k}|^2 |z_{k+1}|^2 +2|z_{k+1}|^2 + |z_{k+2}|^2- 98^2 \\ \ge |z_{k}|^2 |z_{k+1}|^2 - 98^2 $$

So it is enough to prove $$ \max_k |z_{k}|^2 |z_{k+1}|^2 \ge 98^2+2016\\ \max_k |z_{k}| |z_{k+1}| \ge \sqrt{98^2+2016}\\ \max_k |r_{k+1}+r_{k+2} | |r_{k+2}+r_{k} | \ge 4 \sqrt{98^2+2016}\\ $$

Using the general fact that $|r_{k+2}+r_{k} | \ge ||r_{k+2}|-|r_{k}| |$ it is enough to prove $$ \max_k |(|r_{k+1}|-|r_{k+2}| )(|r_{k+2}|-|r_{k}| ) | \ge 4 \sqrt{98^2+2016}\\ $$

On the LHS we have the three values $$ |(|r_{1}|-|r_{2}| )(|r_{2}|-|r_{3}| ) | = |(84-42)(42-28)|= 588\\ |(|r_{2}|-|r_{3}| )(|r_{3}|-|r_{1}| ) | = |(42-28)(28-84)|= 784\\ |(|r_{3}|-|r_{1}| )(|r_{1}|-|r_{2}| ) | = |(28-84)(84-42)|= 2352\\ $$

Hence it is enough to establish $$ 2352 \ge 4 \sqrt{98^2+2016} \simeq 431.2 $$ which is certainly true.

Discussion: Looking again at the task, it is likely that this appeared in the year 2016, hence the RHS was chosen 2016. With regard to the last result, which was again achieved by crude approximations, one can compute that the RHS can be much bigger. In fact, let the RHS be $x$, then the largest $x$ under the given approximations is obtained by $$ 2352 = 4 \sqrt{98^2+x} \\ x = 336140 $$ hence we can rewrite the original task as:

Find some $Q$ such that
$$ \max\{|z_{1}z_{2}|^2-2|z_{1}|^2-|z_{3}|^2,|z_{2}z_{3}|^2-2|z_{2}|^2-|z_{1}|^2,|z_{3}z_{1}|^2-2|z_{3}|^2-|z_{2}|^2\}\}\ge Q \ge 336140 $$ Choosing $Q=336140$ has just been proven, some higher $Q$ might be established with finer approximations, the hardest task would be to find the highest $Q$ for which the statement holds.