TITLE: Rectangles in rectangles and $(b^2-a^2)^2\le (ax-by)^2+(bx-ay)^2$
QUESTION [6 upvotes]: When does an $a\times b$ rectangle fit inside an $x\times y$ rectangle? I have an algebraic condition which I can diagram geometrically, and I'd like a good geometric argument.
Assume $0<a<b$, $0<x<y$. Then the one rectangle fits in the other iff $a\le x$ and either:
$$b\le y$$
or:
$$(b^2-a^2)^2 \le (ax-by)^2+(ay-bx)^2$$
which I found algebraically, with help from Mathematica. One possible diagram is

and then the last condition is equivalent to
$$
|\boldsymbol{\alpha \times \beta}|^2 \le
|\boldsymbol{\alpha \times \gamma}|^2 +
|\boldsymbol{\beta \times \gamma}|^2
$$
This can also be interpreted in terms of the lengths of these three vectors and the sines of the angles between them.
Is there a nice geometric argument from this diagram to fitting the one rectangle inside the other?

REPLY [7 votes]: The sense of this inequality is that it turns out into equality when one of rectangles is inscribed in another as on the picture. 
We get $ax-by=a(m+n)-b(p+q)=am-bp=a^2\cos \varphi-b^2\cos \varphi=(a^2-b^2)\cos \varphi$, analogously $ay-bx=a(p+q)-b(m+n)=aq-bn=(a^2-b^2)\sin \varphi$, and your relation $(a^2-b^2)^2=(ax-by)^2+(ay-bx)^2$ reads as $\cos^2\varphi+\sin^2 \varphi=1$.