# Triangle Inequality/Real Numbers

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## Theorem

Let $x, y \in \R$ be real numbers.

Let $\size x$ denote the absolute value of $x$.

Then:

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

### General Result

Let $x_1, x_2, \dotsc, x_n \in \R$ be real numbers.

Let $\size x$ denote the absolute value of $x$.

Then:

$\ds \size {\sum_{i \mathop = 1}^n x_i} \le \sum_{i \mathop = 1}^n \size {x_i}$

## Proof 1

 $\ds \size {x + y}^2$ $=$ $\ds \paren {x + y}^2$ Square of Real Number is Non-Negative $\ds$ $=$ $\ds x^2 + 2 x y + y^2$ Square of Sum $\ds$ $=$ $\ds \size x^2 + 2 x y + \size y^2$ Square of Real Number is Non-Negative $\ds$ $\le$ $\ds \size x^2 + 2 \size {x y} + \size y^2$ Negative of Absolute Value $\ds$ $=$ $\ds \size x^2 + 2 \size x \cdot \size y + \size y^2$ Absolute Value Function is Completely Multiplicative $\ds$ $=$ $\ds \paren {\size x + \size y}^2$ Square of Sum
$\size {x + y} \le \size x + \size y$

$\blacksquare$

## Proof 2

This can be seen to be a special case of Minkowski's Inequality for Sums, with $n = 1$.

$\blacksquare$

## Proof 3

We have that Real Numbers form Ordered Integral Domain.

Therefore Sum of Absolute Values on Ordered Integral Domain applies directly.

$\blacksquare$

## Proof 4

We do a case analysis.

### $(1): \quad x \ge 0, y \ge 0$

 $\ds x$ $\ge$ $\ds 0$ $\ds y$ $\ge$ $\ds 0$ $\ds \leadsto \ \$ $\ds \size {x + y}$ $=$ $\ds x + y$ $\ds$ $=$ $\ds \size x + \size y$

$\Box$

### $(2): \quad x \le 0, y \le 0$

 $\ds x$ $\le$ $\ds 0$ $\ds y$ $\le$ $\ds 0$ $\ds \leadsto \ \$ $\ds \size {x + y}$ $=$ $\ds -x - y$ $\ds$ $=$ $\ds \size x + \size y$

$\Box$

### $(3): \quad x \ge 0, y \le 0$

We have that $\size x = x$ and $\size y = -y$.

In this case we show:

$\size {x + y} \le \max \set {\size x, \size y}$

Let $\size x \le \size y$.

Then:

 $\ds x$ $\le$ $\ds -y$ $\ds \leadsto \ \$ $\ds y$ $\le$ $\ds y + x$ as $x \ge 0$ $\ds$ $\le$ $\ds 0$ $\ds \leadsto \ \$ $\ds \size {x + y}$ $=$ $\ds -\paren {x + y}$ $\ds$ $\le$ $\ds -y$ $\ds$ $=$ $\ds \size y$

Let $\size x \ge \size y$.

Then:

 $\ds x$ $\ge$ $\ds -y$ $\ds \leadsto \ \$ $\ds x$ $\ge$ $\ds x + y$ as $y \le 0$ $\ds$ $\ge$ $\ds 0$ $\ds \leadsto \ \$ $\ds \size {x + y} =$ $=$ $\ds x + y$ $\ds$ $\le$ $\ds x$ $\ds$ $=$ $\ds \size x$

We have $\max \set {a, b} \le a + b$ for positive real numbers $a$ and $b$.

The result follows by taking $a = \size x$ and $b = \size y$.

$\Box$

### $(4): \quad x \le 0, y \ge 0$

Follows by symmetry from the case $(3)$.

$\blacksquare$

## Proof 5

From Negative of Absolute Value, it is sufficient to prove that:

$\size x + \size y \ge x + y$

and:

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

By definition of absolute value:

$x \le \size x$

and:

$y \le \size y$

Then:

$x + y \le \size x + \size y$

We also have that:

 $\ds -x$ $\le$ $\ds \size {-x}$ $\ds$ $=$ $\ds \size x$ Absolute Value of Negative $\ds -y$ $\le$ $\ds \size {-y}$ $\ds$ $=$ $\ds \size y$ Absolute Value of Negative $\ds \leadsto \ \$ $\ds -\paren {x + y}$ $=$ $\ds \paren {-x} + \paren {-y}$ $\ds$ $\le$ $\ds \size x + \size y$

Hence the result.

$\blacksquare$

## Examples

### Example: $\size {-1 + 3}$

$2 = \size {-1 + 3} \le \size {-1} + \size 3 = 1 + 3 = 4$