Let (M,d) be a metric space. Define a new metric as $\displaystyle p(x,y)=\frac{d(x,y)}{1+d(x,y)}$ and prove that $\displaystyle p(x,z)\leq p(x,y)+p(y,z)$

**Sorry just had a huge breakthrough, this is now solved**

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- Sep 28th 2010, 07:07 PMemathinstructionmetric space triangle inequality
Let (M,d) be a metric space. Define a new metric as $\displaystyle p(x,y)=\frac{d(x,y)}{1+d(x,y)}$ and prove that $\displaystyle p(x,z)\leq p(x,y)+p(y,z)$

**Sorry just had a huge breakthrough, this is now solved**