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I'm referring to this post: How to find the maximum and minimum value of $\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|$ (where $|z_1|=|z_2|=|z_3|=1$)?

(I can't make a comment because I don't have enough rep. yet, and adding a post to it doesn't seem appropriate).

Why does the method in OP not work? Isn't the LHS at most equal to 12 (and so there should exist a case where it does equal to 12)? What condition makes it not equal to 12 in this case?

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    $\begingroup$ The LHS is at most equal to 12, but that doesn't mean that there is a case where it attains 12. $\endgroup$ – Joppy Apr 13 '17 at 12:19
  • $\begingroup$ @Joppy I'm not sure if I understand why, but I feel as if I do. In a normal triangle inequality (where a,b are just elements in the real set), $|a+b| \leq |a| + |b|$ equality if $a = b$. But why does this not apply to the complex numbers? $\endgroup$ – Twenty-six colours Apr 13 '17 at 12:33
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    $\begingroup$ It has equality when $a$ and $b$ are of the same sign, or point at the same direction. The same works for complex numbers, but you can not have $z_1-z_2$, $z_2-z_3$ and $z_3-z_1$ to point at the same direction (unless all of them equal 0). $\endgroup$ – walak Apr 13 '17 at 12:38
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Basically, what OP does, is saying let $a=z_1-z_2$, $b=z_2-z_3$ and $c=z_3-z_1$. Then length of each of $a$, $b$ and $c$ is at most two, so $$|a^2+b^2+c^2|\leq 2^2+2^2+2^2=12.$$

This is totally true, but it does not use the fact, that $a$, $b$ and $c$ are not some random numbers with length less than 2, they have a more strict relationship (for instance one can notice, that $a+b+c=0$). And this gives us the possibility to prove more strict inequality for them.

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  • $\begingroup$ I see, but for a triangle inequality on scalars $a,b \in \mathbb{R}$, we have $|a+b| < |a| + |b|$ iff $a = b$, i.e. max of $|a+b|$ occurs when $a=b$. Why is this not true in our case here (when $a = b = c$)? $\endgroup$ – Twenty-six colours Apr 13 '17 at 12:37
  • $\begingroup$ @PaulWoch I answered on the same question under your original post. $\endgroup$ – walak Apr 13 '17 at 12:53
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Triangle inequality did not "fail," the bound in OP is correct, just not efficient. The triangle inequality applies, generally speaking, to two points. But to simultaneously maximize the distance between three points, you might need to make $z_1$ closer to $z_2$ so that you can make $z_3$ further from both. In this case, I believe to maximize $|\sum_{cyc}(z_a - z_b)^2|$ you'd pick the $z_a$ to form an equilateral polygon on the unit circle.

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The statement that a variable quantity is "at most equal to $12$" is another way of saying that the quantity is "less than or equal to $12$". That number $12$ only serves as an upper bound. The statement carries no guarantee that there is a case where it does equal $12$.

Perhaps this example makes it clearer: $x^2 \ge -1$ for all real numbers $1$. This is a true statement, which I'm sure you understand. However, there is no guarantee that there is a case where $x^2$ does equal $-1$; an in fact, $x^2$ never equals $-1$.

You may wonder: "Why bother writing $x^2 \ge -1$ when you can write a stronger statement $x^2 > -1$? Or an even still stronger statement like $x^2 \ge 0$, which is as strong as possible?"

Writing strong and efficient inequalities is one of the more important tasks in mathematics. Sometimes it's easy to make the inequality more efficient (as it is for the example $x^2 \ge -1$), sometimes it's quite hard. In fact, writing the "most efficient inequality" is more or less the same as determining the maximal or minimal value of an expression, and this is a hard task in general. Entire chapters in Calculus books are devoted to this subject.

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