Reverse Triangle Inequality/Real and Complex Fields

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Theorem

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

Then:

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

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


Corollary 1

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.


Corollary 2

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.


Corollary 3

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

Then:

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

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


Proof 1

Let $X$ denote either $\R$ or $\C$ as appropriate.

From Real Number Line is Metric Space and Complex Plane is Metric Space the distance function $d: X \times X \to \R$ can be defined as:

$\map d {x, y} = \size {x - y}$


From the Reverse Triangle Inequality as applied to metric spaces:

$(1): \quad \forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$


Let $z = 0$.

Then $(1)$ translates to:

$\forall x, y, z \in X: \size {\size {x - 0} - \size {y - 0} } \le \size {x - y}$

Hence the result.

$\blacksquare$


Proof 2

From proof $2$ of corollary $1$ to this result, which is derived independently:

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


There are two cases:

$(1): \quad \size x \ge \size y$

We have :

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

and the proof is finished.

$\Box$


$(2): \quad \size y \ge \size x$

We have:

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

But:

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

and:

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


From this we have:

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

Since, by Negative of Absolute Value, we have that:

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

it follows that:

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

The result follows.

$\blacksquare$


Examples

Example: $\size {6 - \paren {-1} }$

$7 = \size {6 - \paren {-1} } \ge \size 6 - \size {-1} = 6 - 1 = 5$


Sources

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