Triangle Inequality/Real Numbers/Proof 1

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Theorem

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

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


Then:

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


Proof

\(\ds \size {x + y}^2\) \(=\) \(\ds \paren {x + y}^2\)
\(\ds \) \(=\) \(\ds x^2 + 2 x y + y^2\)
\(\ds \) \(=\) \(\ds \size x^2 + 2 x y + \size y^2\)
\(\ds \) \(\le\) \(\ds \size x^2 + 2 \size {x y} + \size y^2\) Negative of Absolute Valueā€ˇ
\(\ds \) \(=\) \(\ds \size x^2 + 2 \size x \cdot \size y + \size y^2\) Absolute Value of Product
\(\ds \) \(=\) \(\ds \paren {\size x + \size y}^2\)


Then by Order is Preserved on Positive Reals by Squaring:

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

$\blacksquare$


Sources