Metric space problem (Triangle inequality)

Question

Let $X = \mathbb{R}^+$

$f\colon X\times X \to [0,\infty)$

(Triangle inequality)

$$\frac{|x-y|}{(1+x)(1+y)}\leq\frac{|x-z|}{(1+x)(1+z)}+\frac{|z-y|}{(1+z)(1+y)}$$

I only known than $|x-y|=|x-z+z-y|=|(x-z)+(z-y)|\leq|x-z|+|z-y| $

Answer 1

We need to prove that $$|x-z|(1+y)+|y-z|(1+x)\geq|x-y|(1+z)$$ or $$\left(|x-z|(1+y)+|y-z|(1+x)\right)^2\geq(x-y)^2(1+z)^2$$ or $$(x-z)^2(1+y)^2+2(1+x)(1+y)|(x-z)(y-z)|+(y-z)^2(1+x)^2\geq(x-y)^2(1+z)^2.$$ Now, $$(x-z)^2(1+y)^2+(y-z)^2(1+x)^2-(x-y)^2(1+z)^2=$$ $$=(x-z)^2(1+y)^2+((y-z)(1+x)-(x-y)(1+z))((y-z)(1+x)+(x-y)(1+z))=$$ $$=(x-z)^2(1+y)^2+(2y-x-z+xy+yz-2xz)(xy+x-z-yz)=$$ $$=(x-z)^2(1+y)^2+(2y-x-z+xy+yz-2xz)(1+y)(x-z)=$$ $$=(1+y)(x-z)((x-z)(1+y)+2y-x-z+xy+yz-2xz)=$$ $$=(1+y)(x-z)(x+xy-z-yz+2y-x-z+xy+yz-2xz)=$$ $$=(1+y)(x-z)(2xy-2xz-2z+2y)=2(1+x)(1+y)(x-z)(y-z).$$ Thus, we need to prove that: $$(1+x)(1+y)|(x-z)(y-z)|+(1+x)(1+y)(x-z)(y-z)\geq0,$$ which is obvious.