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If $\displaystyle |z_1| = |z_2| = |z_3| = R$ then prove $\displaystyle \sum_{z_1, z_2, z_3} |z_1 - z_2|\cdot|z_2 - z_3| \le 9R^2$ for $z_1,z_2,z_3 \in \mathbb{C}$.

$$\sum_{z_1, z_2, z_3} |z_1 - z_2|\cdot|z_2 - z_3| = \sum_{z_1, z_2, z_3} |z_1z_2 - z_1z_3-z_2^2 + z_3z_2| \le \sum_{z_1, z_2, z_3} |z_1z_2| + |z_1z_3|+|z_2^2| + |z_3z_2| = 12R^2$$

$$\sum_{z_1, z_2, z_3} |z_1 - z_2|\cdot|z_2 - z_3| \le 12R^2$$

Therefore $\displaystyle\sum_{z_1, z_2, z_3} |z_1 - z_2|\cdot|z_2 - z_3| = 12R^2$ for some $z_1, z_2, z_3$. Hence the given question is incorrect.

  • I don't think the question is incorrect instead I think I have commited some blunder that I am unable to spot. Where did I go wrong ?
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    But that doesn't mean the question is incorrect. It just means that you can do better in bounding that expression. I mean, it is bounded by $12R^2$, but other calculation could show it's in fact bounded by $9R^2$. – 2017-02-15
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    @flytothesurface I get that but Triangle inequality has a equality so for some $z_1, z_2,z_3$ the equality if $12R^2$ will be achived, but that violates the question. – 2017-02-15
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    Hm, I think you're right. To be sure of that, it would be good to find some values of $z_1$, $z_2$ and $z_3$ satisfying the equality. – 2017-02-15
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    @flytothesurface I don't know Mathematica/Matlab or things like that. – 2017-02-15
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    Hint: geometrically, let $a,b,c$ be the sides of the triangle $z_1 z_2 z_3\,$, then the inequality follows from: $ab+bc+ca \le a^2+b^2+c^2 \le 9 R^2\,$. The first of the two inequalities is easy to prove algebraically, the second one is a classic related to $OH^2$ in a triangle, see [here](http://math.stackexchange.com/questions/939902/alternate-proof-for-a2b2c2-le-9r2) for example. – 2017-02-16
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    @dxiv Proving that inequality is not the main concern rather why I got bad bound with triangle inequality ? – 2017-02-16
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    @A---B That's not a `bad` bound. $12$ is a correct upper bound, it's just not the lowest upper bound. Your only mistake is stating that "*therefore ...$=12 R^2$ for some $z_1,z_2,z_3$*". That does not follow from what you proved and, in fact, there exist no $z_1,z_2,z_3$ such that the equality $=12R^2$ holds. – 2017-02-16
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    @dxiv Correct me if I am wrong the triangle inequality is $|z\pm z_1| \le |z| + |z_1|$, so did not it follow directly from triangle inequality statement that $|z\pm z_1| = |z| + |z_1|$ for some $z_1, z$. I did the same in my proof. Then why I have to prove $|z\pm z_1| = |z| + |z_1|$ ? – 2017-02-16
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    @dxiv If I prove that then would not that means Triangle inequality statement is false one. – 2017-02-16
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    @A---B The triangle inequality does not guarantee that the equality case is possible in general. Take for example $b=-a\,$, then by the triangle inequality $|a+b|\le|a|+|b|= 2 |a|\,$. This is correct, of course, and $2 |a|$ is a valid upper bound, but the lowest upper bound is of course $0$ since $a+b=0\,$. – 2017-02-16
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    @dxiv Then why we write $\le$ ? why that don't make the statement incorrect ? – 2017-02-16
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    @A---B Sorry, not sure where you see a problem. For example $9 \le 12$ is a correct statement. – 2017-02-16
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    @dxiv I did not know that $9 \le 12$ is valid I always thought $9 \lt 12$ is the valid statement. Sure this was a big blunder, thanks. – 2017-02-17

1 Answers 1

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Let $|z_1-z_2|=a$, $|z_2-z_3|=b$ and $|z_3-z_1|=c$.

Hence, $a+b=|z_1-z_2|+|z_2-z_3|\geq |z_1-z_2+z_2-z_3=z_1-z_3|=c$ and

we need to prove that $ab+ac+bc\leq9R^2$ for all $\Delta ABC$.

Indeed, let $a^2=x$, $b^2=y$ and $c^2=z$.

Hence, since $ab+ac+bc\leq a^2+b^2+c^2=x+y+z$, we need to prove that $$x+y+z\leq\frac{9a^2b^2c^2}{16S^2}$$ or $$x+y+z\leq\frac{9xyz}{\sum\limits_{cyc}(2xy-x^2)}$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.

Done!

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    Did you read my comments below the question ? – 2017-02-15
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    Please just read the second comment under the question. – 2017-02-15
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    If you are correct then that means equality of "Triangle inequality" does not always hold ? – 2017-02-15
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    Oh I get it. The equality will hold for $z_1 = z_2 = z_3 = 0$. So no contradiction. Am I correct ? – 2017-02-15
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    I did not understand, How ? – 2017-02-15
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    @A---B The equality occurs for $z_1$, $z_2$ and $z_3$ are roots of the equation $z^3=R^3$, id est, where our triangle is equilateral. – 2017-02-16
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    But that does not answer my question of does equality of triangle inequality always hold or not ? – 2017-02-16
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    @A---B If $O$ is an origin so $O$ is a circumcenter of $\Delta ABC$. The equality occurs when $\Delta ABC$ is equilateral. – 2017-02-16
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    Ok I get that. When $\triangle ABC$ is equilateral I will $9R^2$ on LHS **but** in my proof by triangle inquality I got $12R^2$ as the maxima. So something should be wrong here. – 2017-02-16
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    @A---B $9R^2<12R^2$. When does the equality occur in your inequality? Take $R>0$. – 2017-02-16
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    Yes. As you said equality never occur in my inequality but I used triangle inequlity which is a **slack inequality**, so it has a equality ($12R^2$ in this case). But in this case that equality is never fulfilled (maxima as you proved is $9R^2$) so it means that "triangle inequality's equality" does not always hold. right ? – 2017-02-16
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53760/discussion-between-michael-rozenberg-and-a-b). – 2017-02-16