$x,y,z$ $\in\mathbb R$ then $|x-y| \leq|x-z|+|y-z|$
Prove this statement.
I thought it was the triangle inequality, but I can't seem to end up with the correct order.
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It is the triangle inequality. Try to draw three labeled points $x,y,z$ and lines between them. – Jul 16 '14 at 15:33
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Note that: $|x-y|=|(x-z)+(z-y)|$, now use the triangle inequality and the fact that $|z-y|=|y-z|$. – DiegoMath Jul 16 '14 at 15:35
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See Proving the inequality $|a-b| \leq |a-c| + |c-b|$ for real $a,b,c$ or Proof for absolute value inequality of three variables: $|x-z| \leq |x-y|+|y-z|$. – Martin Sleziak Jan 19 '17 at 02:01
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By the triangle inequality for any $x,y,z \in \mathbb{R}$, $$|x-y| = |x-z+z-y|= |(x-z)+(z-y)| \leq |x-z|+|z-y|=|x-z|+|y-z|$$
Surb
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