Reverse Triangle Inequality

Assuming $0\le x\le y$ $||x|-|y||\le |x-y|$

Proof

Assume we already proved laws 1-5 for Absolute Function. and the Triangle Inequality Understand aswell, that to show an absolute function is true. you need to show $|a|\le b\iff -b\le a\le b$ Let $x,y\in R$

$|x|-|y|\le |x-y|$ let $m=x-y$ and $n=y$ $|m+n|\le |m|+|n|$ by Triangle Inequality $|x|\le |x-y|+|y|$ $|x|-|y|\le |x-y|$ 2. $|x|-|y|\ge -|x-y|$ $m=y-x$ $n=x$ $|m+n|\le |m|+|n|$ by Triangle Inequality $|y|\le |y-x|+|x|$ $-|y-x|\le |x|-|y|$

Since both the propositions in $|a|\le b\iff -b\le a\le b$ are shown, we have proved the inequality.