Triangle Inequality

Let $x,y\in R$ Assuming $0\le x\le y$

Then, $|x+y|\le |x|+|y|$

Proof

Assume we already proved laws 1-5 for Absolute Function. Let $x,y\in R$ Consider two cases. Case 1: Assume $x+y>0$ Then $|x+y|=+(x+y)$ (by definition of Absolute Function) We know that $x\le |x|$ and $y\le |y|$ so it follows that $x+y\le |x|+|y|$ Case 2: Assume $x-y<0$ Then, $|x+y|=-(x+y)=-x-y$ (by piecewise definition of Absolute Function) We know that $-x\le |x|$ and $-y\le |y|$ so it follows that $-x-y\le |x|+|y|$ Since both cases of the absolute function are fufilled, we can conclude that: $|x+y|\le |x|+|y|$