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

Question

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 $z_1,z_2,z_3 \in \mathbb{C}$.

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

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

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.

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• 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 ?

Answer 1

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!