# 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.)

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

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

### Proofs

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 $a=BC, b=AC, c=AB.$

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:

(1)

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

and

(2)

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

From (1) we get

(1')

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

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

(2')

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

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

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

or,

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

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

$e^{i\alpha}+e^{i\beta}+e^{i\gamma}=0.$

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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