Triangle Inequality/Real Numbers/General Result

Theorem
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
Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * $\ds \size {\sum_{i \mathop = 1}^n x_i} \le \sum_{i \mathop = 1}^n \size {x_i}$

$\map P 1$ is true by definition of the usual ordering on real numbers:
 * $\size {x_1} \le \size {x_1}$

Basis for the Induction
$\map P 2$ is the case:
 * $\size {x_1 + x_2} \le \size {x_1} + \size {x_2}$

which has been proved in Triangle Inequality for Real Numbers.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\ds \size {\sum_{i \mathop = 1}^k x_i} \le \sum_{i \mathop = 1}^k \size {x_i}$

Then we need to show:
 * $\ds \size {\sum_{i \mathop = 1}^{k + 1} x_i} \le \sum_{i \mathop = 1}^{k + 1} \size {x_i}$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Also see

 * Triangle Inequality for Indexed Summations