Prove $d(\vec{x} , \vec{z})\leq d(\vec{x},\vec{y})+d(\vec{y},\vec{z})$

Question

Prove $d(\vec{x} , \vec{z})\leq d(\vec{x},\vec{y})+d(\vec{y},\vec{z})$.

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My attempt:

$$d(\vec {x}, \vec {z})= ||\vec{z}-\vec{x}||$$ $$=||\vec{z}-\vec{y}+\vec{y}-\vec{x}|| = ||(\vec{z}-\vec{y})+(\vec{y}-\vec{x})||$$

so we can re-write the problem as,

$$||(\vec{z}-\vec{y})+(\vec{y}-\vec{x})|| \leq ||\vec{y}-\vec{x}||+||\vec{z}-\vec{y}||$$

If $\vec{y}=\vec{x}$ then

$$||(\vec{z}-\vec{y})+(\vec{y}-\vec{y})|| \leq ||\vec{y}-\vec{y}||+||\vec{z}-\vec{y}||$$

Implies

$$||\vec{z}-\vec{y}||=||\vec{z}-\vec{y}||$$ $$d(\vec{y},\vec{z})= d(\vec{y},\vec{z})$$

This accounts for the equal part of the inequality "$\leq$" I am not sure how to prove the "less than" part.

Answer 1

Use the triangle inequality. $$||\vec{x}+\vec{y}|| \le ||\vec{x}||+||\vec{y}||$$ It follows that $$||\vec{z}-\vec{y}+\vec{y}-\vec{x}|| = ||(\vec{z}-\vec{y})+(\vec{y}-\vec{x})|| \le ||\vec{y}-\vec{x}||+||\vec{z}-\vec{y}||$$