Reverse Triangle Inequality/Real and Complex Fields/Corollary

Theorem

Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.

Then:

$\size {x - y} \ge \size x - \size y$

where $\size x$ denotes either the absolute value of a real number or the complex modulus of a complex number.

Proof 1

From the Reverse Triangle Inequality:

$\cmod {x - y} \ge \cmod {\cmod x - \cmod y}$

By the definition of both absolute value and complex modulus:

$\cmod {\cmod x - \cmod y} \ge 0$

As:

$\cmod x - \cmod y = \pm \cmod {\cmod x - \cmod y}$

it follows that:

$\cmod {\cmod x - \cmod y} \ge \cmod x - \cmod y$

Hence the result.

$\blacksquare$

Proof 2

By the Triangle Inequality:

$\cmod {x + y} - \cmod y \le \cmod x$

Let $z = x + y$.

Then $x = z - y$ and so:

$\cmod z - \cmod y \le \cmod {z - y}$

Renaming variables as appropriate gives:

$\cmod {x - y} \ge \cmod x - \cmod y$

$\blacksquare$

Proof 3

Let $z_1$ and $z_2$ be represented by the points $A$ and $B$ respectively in the complex plane.

From Geometrical Interpretation of Complex Subtraction, we can construct the parallelogram $OACB$ where:

$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
$BA$ represents $z_1 - z_2$.

But $OA$, $OB$ and $BA$ form the sides of a triangle.

The result then follows directly from Sum of Two Sides of Triangle Greater than Third Side.

$\blacksquare$