Triangle Inequality/Real Numbers/Proof 5

Theorem

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

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

Then:

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

Proof

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.

$\blacksquare$