# Chaitanya's Random Pages

## April 22, 2013

### A few geometric inequalities proved using complex numbers

Filed under: mathematics — ckrao @ 11:33 am

The following inequalities all follow from the elementary triangle inequality for complex numbers:

$\displaystyle \left|z_1 \right| + \left| z_2 \right| \geq \left| z_1 + z_2 \right|.$

Equality holds if and only if $0, z_1, z_2$ are collinear or in other words $z_2/z_1$ is a real number when $z_1 \neq 0$.

\begin{aligned} (p-q)(r-s) + (p-s)(q-r) &= pr - qr - ps + qs + pq - qs - pr + rs\\ &= pq - qr - ps + rs\\ &= (p-r)(q-s) \end{aligned}

Hence

\begin{aligned}\left| p-r \right| \left| q-s\right| &= \left| (p-r)(q-s) \right| \\ &= \left| (p-q)(r-s) + (p-s)(q-r)\right| \\&\leq \left| (p-q)(r-s)\right| + \left| (p-s)(q-r)\right|\end{aligned}

Hence if $P, Q, R, S$ are four points in the plane,

$\displaystyle PR \times QS \leq PQ \times RS + PS \times QR, \quad \quad \quad (1)$

which is Ptolemy’s inequality. Equality holds if and only if

$\displaystyle \frac{(p-s)(q-r)}{(p-q)(r-s)} \in \mathbb{R}$. The ratio $(p-s)/(p-q)$ is a complex number with argument $\angle QPS$ while the ratio $(q-r)/(r-s)$ has argument $\pi - \angle SRQ$. Hence for the product of these ratios to be real means $\angle QPS + \angle SRQ$ is a multiple of $\pi$. In other words, the points $P, Q, R, S$ lie on a circle.

\begin{aligned}\frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)} &= \frac{bc(c-b) + ca(a-c) + ab(b-a)}{(a-b)(b-c)(c-a)} \\ &= 1 \end{aligned}

From this,

\begin{aligned}\frac{ |b| |c|}{|a-b| |a-c|} + \frac{|c| |a|}{|b-c| |b-a|} + \frac{|a| |b|}{|c-a| |c-b|} &\geq \left| \frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)}\right|\\ &= 1. \end{aligned}

In other words, if $P, A, B, C$ are four points in the plane, $(PB\times PC)/(AB \times AC) + (PC \times PA)/(BC \times AB) + (PA \times PB)/(AC \times BC) \geq 1$, or

$\displaystyle (BC \times PB \times PC) + (CA \times PC \times PA) + (AB \times PA \times PB) \geq AB \times BC \times CA.\quad (2)$

3. Similarly we have

\begin{aligned} a^2(b-c) + b^2(c-a) + c^2(a-b) &= -(a-b)(b-c)(c-a), \end{aligned}

from which

$\displaystyle PA^2 \times BC + PB^2 \times CA + PC^2 \times AB \geq AB \times BC \times CA.\quad \quad \quad (3)$

4. Finally, we have the equality

\begin{aligned} & a^3(b-c) + b^3(c-a) + c^3(a-b) \\&= (a + b + c)(a^2(b-c) + b^2(c-a) + c^2(a-b)) - \left[ (b+c) a^2(b-c) + (c + a) b^2 (c-a) + (a+b)c^2(a-b) \right]\\ &= -(a+b+c)(a-b)(b-c)(c-a) + \left[ a^2(b^2-c^2) + b^2(c^2 - a^2) + c^2(a^2-b^2) \right] \quad \text{(from the equality in 3.)}\\ &= -(a-b)(b-c)(c-a)(a + b + c)\end{aligned}

from which

$\displaystyle PA^3 \times BC + PB^3 \times CA + PC^3 \times AB \geq 3 AB \times BC \times CA \times PG, \quad \quad \quad (4)$

where $G$ is the centroid of $\triangle ABC$ (i.e. the vector sum $PA + PB + PC$ is $3PG$).

See [2] for more details of special cases of these inequalities.

#### References

[1] A. Bogomolny, Complex Numbers and Geometry from Interactive Mathematics Miscellany and Puzzles  http://www.cut-the-knot.org/arithmetic/algebra/ComplexNumbersGeometry.shtml#cycl, Accessed 22 April 2013.

[2] T. Andreescu and D. Andrica, Proving some geometric inequalities by using complex numbers, Educatia Matematica Vol. 1, Nr. 2 (2005), pp. 19–26.