Proving Inequality: d(x,y) = d1(x,y)/[1+d1(x,y)] as a Valid Distance in R^n

[RenOdur]

Homework Statement

the actual problem is to show that d(x,y)=d1(x,y)/[1+d1(x,y)] expresses a distance in R^n if d1(x,y) is a distance in R^n.Based on theory I have to show that
i) d(x,y)>=0 ,
ii)d(x,y)=d(y,x) and
iii)d(x,y)<= d(x,z)+d(z,y)
i've proven the first two so basically how can i get from d1(x,y)<= d1(x,z)+d1(z,y)(the above three statements apply for d1(x,y) since d1(x,y) is already a distance in R^n ) to d1(x,y)/[1+d1(x,y)]<=d1(x,z)/[1+d1(x,z)]+d1(z,y)/[1+d1(z,y)] ?

 

[K29]

hint:Divide through all of (iii) by 1+ d1(x,y). You should then be able to write another inequality using what you know in (iii) and (i) on the denominator

 

[RenOdur]

hint:Divide through all of (iii) by 1+ d1(x,y). You should then be able to write another inequality using what you know in (iii) and (i) on the denominator

Let me get this right.do you suggest that i divide d(x,y)<= d(x,z)+d(z,y) by 1+d1(x,y) in order to reach an inequality that is true?or do you suggest dividing d1(x,y)<= d1(x,z)+d1(z,y) by 1+d1(x,y) in order to reach d(x,y)<= d(x,z)+d(z,y)?

 

[K29]

well first you need to write (iii) in terms of d1 (which is also a distance)
then after noticing some things you can easily get to
d1(x,y)/[1+d1(x,y)]<=d1(x,z)/[1+d1(x,z)]+d1(z,y)/[1+d1(z,y)]