Q. Prove that |d(x,y)-d(y,z)|≤d(x,y)

Question

This question is about Functional analysis.

Answer 1

When y=z, d(y,z)=0 because this is the distance between y and itself or z and itself.

Therefore |d(x,y)-d(y,z)|=|d(x,y)-0|=d(x,y). The equality is included in the inequality ≤.

When y≠z, d(y,z)≠0 and d(y,z)>0 by definition of d. Also, d(x,y)>0.

If d(x,y)=d(y,x)=aᵣ>0 and d(y,z)=d(z,y)=bᵣ>0; -bᵣ<0, add aᵣ to each side: aᵣ-bᵣ<aᵣ.

Therefore |d(x,y)-d(y,z)|=|aᵣ-bᵣ|<|aᵣ|. But |aᵣ|=d(x,y) so |d(x,y)-d(y,z)|<aᵣ.

Combining the two conditions: |d(x,y)-d(y,z)|≤aᵣ QED