2
$\begingroup$

Let $\|\cdot\|$ be a norm on a vector space $V$, and define, for each ordered pair of vectors, the scalar $d(x,y) = \|x-y\|$, called distance between $x$ and $y$. Prove the following results for all $x,y,z\in V$.

$d(x,y) \leq d(x,z)+d(z,y)$

I tried to prove this in different ways, I tried to prove $\|x-y\|^{2}\leq (||x-z|| + ||z-y||)^{2}$ and expanded everything out, move this and that around, but none of my tactics work. Please give me some insight or trick of how to prove this. Thank you.

  • 0
    Inner products have nothing to do with your work. Is better you say a metric on V2011-10-31

1 Answers 1

1

If you have the axioms of a norm in hand, then as first comment states, it follows from $|a + b| \leq |a| + |b|$. Here $a$ is the displacement from $x$ to $y$ and $b$ the displacement from $y$ to $z$.

If the norm is defined from the inner product and you want to check that $|a+b| \leq |a| + |b|$, the square of that inequality is equivalent to the Cauchy-Schwarz inequality $(a . b) \leq |a||b|$.