There is a well-known construction that shows that in a right triangle with sides $a\le b<c$, we have $a^2+b^2=c^2$. Inscribe a square of side $c$ into a square of side $a+b$ so that each vertex of the inner square partitions a side of the outer square into segments of lengths $a$ and $b$ in that order, say, clockwise.
I am curious if there is a similarly nice geometric proof that shows the following: if $a\le b<c$ are sides of a right triangle, then so are $2a+b+2c\le a+2b+2c<2a+2b+3c$.
For example, applying this transformation to $(3,4,5)$, we get $(20,21,29)$. It would also be nice if this construction worked even if some sides were allowed to be interpreted as having negative lengths: say, $(-3,4,5)$ yields $(8,15,17)$, and $(-4,3,5)$ yields $(5,12,13)$.