Prove that if x,y,z ∈ R, then |x − y| ≤ |x − z| + |z − y|.
I know that in order to get the inequality I start off with |x-y| = |(x-z)+(z-y)| and my final result should be |x-y| ≤ |(x-z)+(z-y)| using the triangle inequality in the last step. Can someone please help me fill in the missing steps that I need because I don't understand if I'm supposed to use the triangle inequality throughout the steps or if it comes in only at the end.
The triangle inequality allows you to write $|a+b|\leq |a|+|b|$ for all $a$ and $b$ in $\mathbb{R}$, so here you want to apply it with $a=(x-z)$ and $b=(z-y)$. Namely, you establish your inequality through the steps
$$|x-y|=|x-z+z-y|\leq|x-z|+|z-y|.$$
Prove that if x,y,z ∈ R, then |x − y| ≤ |x − z| + |z − y|.
I know that in order to get the inequality I start off with |x-y| = |(x-z)+(z-y)| and my final result should be |x-y| ≤ |(x-z)+(z-y)| using the triangle inequality in the last step.
You are so close! You get to |(x-z)+(z-y)| by adding and subtracting z from the inside the absolute value signs of |x-y|. From there use substitution to let u= x-z and v=z-y, just so it is easier to visualize. Then use the triangle inequality to get |x-y| ≤ |u|+|v|, substitute back to get |x − y| ≤ |x − z| + |z − y| and you are done!
