<s>[INST] Normalized form of Triangule Inequality Extension 
How I can prove $$\mathit{\frac{\left\vert a-b\right\vert}{a+b}\leq\frac{\left\vert a-c\right\vert}{a+c}+\frac{\left\vert c-b\right\vert}{c+b}}$$ where $ a,b,c \in \mathbb N?$

 [/INST] This nice inequality holds for non-negative REAL numbers $a,b,c$.
Suppose $a\ge b$. 
Consider three cases:


*

*$a\ge c \ge b$: we have $$RHS = \frac{a-c}{a+c}+\frac{c-b}{c+b}=\frac{2c(a-b)}{(a+c)(c+b)}.$$ It suffices to show that $$\frac{2c}{(a+c)(c+b)} \ge \frac{1}{a+b},$$ which is equivalent to $2c(a+b) \ge (a+c)(c+b) \Leftrightarrow (a-c)(c-b)\ge 0$, clearly true.

*$c\ge a \ge b$: in this case we have $$\frac{a-b}{a+b} \le \frac{c-b}{c+b}$$
(equivalent to $(a-b)(c+b)\le (a+b)(c-b) \Leftrightarrow 2b(c-a)\ge 0)$, and thus the original inequality is true.

*$a \ge b \ge c$: in this case, similarly we have $$\frac{a-b}{a+b} \le \frac{a-c}{a+c}.$$


We are done.
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