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Let $V$ be an inner product space. The distance between two vectors $\alpha$ and $\beta$ in $V$ is defined by:

$d(\alpha, \beta) = ||\alpha - \beta||$

Show that $d(\alpha, \beta) \leq d(\alpha, \gamma) + d(\gamma, \beta)$.


This one has stumped me. I tried bringing in some sums to help, but they aren't very helpful:

$d(\alpha, \gamma) + d(\gamma, \beta) = \displaystyle\sqrt{\sum\limits_{i=1}^n (\alpha_{i} - \gamma_{i})} + \displaystyle\sqrt{\sum\limits_{i=1}^n (\gamma_{i} - \beta_{i})} \leq \displaystyle\sqrt{\sum\limits_{i=1}^n (\alpha_{i} - \beta_{i})}$

Squaring both sides would just make this thing look even worse...

Visually, I can see that it's true (in $\mathbb{R}^2$ it's just the Triangle Inequality). Could anyone give me a hint as to how I could prove this?

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2 Answers

up vote 4 down vote accepted

Hint: Look at $d(\alpha,\beta)^2 = (||(\alpha - \gamma) + (\gamma - \beta)||)^2$ and use the Cauchy-Schwarz inequality and definition of norm on an inner product space.

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Hint: Use the Cauchy-Schwarz inequality $|\langle x, y\rangle|^2 \leq \langle x, x\rangle\langle y, y \rangle$ to show that $\|x+y\|^2 = \langle x + y, x+y\rangle \leq (\|x\| + \|y\|)^2$ by expanding the inner product.

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