An Inequality Involving Fermat Point

The following problem has been suggested by Leo Giugiuc at the CutTheKnotMath facebook page. (This appears to be Problem 2 that was offered at the 2008 Romanian National Olympiad.)

An Inequality Involving Fermat Point


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Copyright © 1996-2018 Alexander Bogomolny

Complex numbers $a,b,c$ satisfy $a|bc|+b|ca|+c|ab|=0.$ Prove that

$|a-b|\cdot |b-c|\cdot |c-a|\ge 3\sqrt{3}|abc|.$


It is common to employ algebra and complex numbers to solve planar problems. Below, I'll reverse the action by interpreting the problem in geometric terms and solving the result as a problem in triangle geometry.

First note that the condition $a|bc|+b|ca|+c|ab|=0$ tells us that the angles between any two of the three vectors joining the origin to the points $A,B,C$ associated with complex numbers $a,b,c$ are all $120^{\circ}.$ Indeed, the lengths of the the three vectors $a|bc|,b|ca|,c|ab|$ are equal so that, since their sum is $0,$ they form an equilateral triangle.

We shall thus consider $\Delta ABC$ whose Fermat point $F$ lies at the origin. I shall change the notations: let $x=FA, y=FB, z=FC$ and $a=BC, b=AC, c=AB.$

Fermat point

The inequality in question becomes


$abc\ge 3\sqrt{3}xyz.$

To prove the latter inequality, I'll compute the area $S$ of $\Delta ABC$ in two ways:


$S=\frac{1}{2}ab\sin\gamma =\frac{1}{2}bc\sin\alpha =\frac{1}{2}ca\sin\beta$



$\begin{align}\displaystyle S &=\frac{1}{2}xy\sin 120^{\circ}+\frac{1}{2}yz\sin 120^{\circ}+\frac{1}{2}zx\sin 120^{\circ}\\ &=\frac{1}{2}\cos 30^{\circ}(xy+yz+zx)\\ &=\frac{\sqrt{3}}{4}(xy+yz+zx). \end{align}$

From (1) we get


$\displaystyle S^{3}=\frac{(abc)^{2}}{8} \sin\alpha \sin\beta \sin\gamma$

while to (2) I apply the Arithmetic Mean - Geometric Mean:


$\begin{align}\displaystyle S &= \frac{\sqrt{3}}{4}(xy+yz+zx)\ge 3\frac{\sqrt{3}}{4}\sqrt[3]{xy\cdot yz\cdot zx}\\ &=\frac{3\sqrt{3}}{4}\sqrt[3]{xy\cdot yz\cdot zx}. \end{align}$

so that $S^{3}\ge \bigg(\frac{3\sqrt{3}}{4}\bigg)^{3}(xyz)^{2}.$ Combining (1') and (2') shows that

$\displaystyle \frac{(abc)^{2}}{8}\sin\gamma \sin\alpha \sin\beta\ge \bigg(\frac{3\sqrt{3}}{4}\bigg)^{3}(xyz)^{2}.$

For one more step, recollect that in any triangle,

$\displaystyle \sin\alpha\sin\beta\sin\gamma\le\frac{3\sqrt{3}}{8}$

which leads to

$\displaystyle \frac{(abc)^{2}}{8}\frac{3\sqrt{3}}{8}\ge \bigg(\frac{3\sqrt{3}}{4}\bigg)^{3}(xyz)^{2}$


$\displaystyle (abc)^{2}\ge (3\sqrt{3})^{2}(xyz)^{2},$

i.e., (*). Quite obviously, the equality only holds for the equilateral triangle.

Leo Giugiuc has offered a shorter proof which was also suggested in the comments below. After establishing that the angles between the vectors $a,b,c$ (back to his original notations) are $120^{\circ},$ he observes that, say, $|a-b|$ is the third side of the triangle with sides $|a|$ and $|b|,$ opposite the $120^{\circ}$ angle. By the Cosine Law,

$|a-b|^{2}=|a|^{2}+|ab|+|b|^{2}\ge 3|ab|$

because $(|a|-|b|)^{2}=|a|^{2}-2|ab|+|b|^{2}.$ It follows that $|a-b|\ge \sqrt{3}\sqrt{|ab|}.$ Similar inequalities hold for $|b-c|$ and $|c-a|;$ the product of the three amounts to the required inequality.

Amit Itagi offered a variant:

When one od $a,b,c\,$ is $0\,$ the solution is trivial. Otherwise, let $a=|a|e^{i\alpha}$, $b=|b|e^{i\beta}$, $c=|c|e^{i\gamma}$. The constraint implies


Three unit complex numbers summing to $0$ means that the unsigned angle between any two of $\{a,b,c\}$ is $2\pi/3$. Thus, applying cosine rule to the triangle formed by $a$, $b$, and $a-b$,

$\displaystyle\begin{align} &\cos\frac{2\pi}{3}=-\frac{1}{2}=\frac{|a|^2+|b|^2-|a-b|^2}{2|a||b|} \\ &|a-b|^2=|a|^2+|b|^2+|a||b|\geq3(|a||b|)~\text{(AM-GM)}. \end{align}$

Multiplying the cyclical variants together and taking the square root,

$|a-b||b-c||c-a|\geq 3\sqrt{3}|a||b||c|.$

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Copyright © 1996-2018 Alexander Bogomolny