Proof by cases inequality with abs. value and min max

Question

Help me to figure out how to make a proof by cases of the following statement:

For all $x, y \in \mathbb{R}$ if $| x − y | < \min(x, y)$ then $\max(x, y) < 2 \min(x, y).$

A proof by cases is of the schema:

case 1 : ...

sub-case a: ...

case 2: ...

Answer 1

First case $\min(x,y)=x$:

Then $\max(x,y)=y$ and $\max(x,y)<2\min(x,y)$ becomes to $y<2x$. From $|x-y|<\min(x,y)=x$ we have $-x < x-y < x$, the left part says $y<2x$.

Second case $\min(x,y)=y$:

Then $\max(x,y)=x$ and $\max(x,y)<2\min(x,y)$ becomes to $x<2y$. From $|x-y|<\min(x,y)=y$ we have $-y < x-y < y$, the right part says $x<2y$.