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?