A - Vertex at bottom left
B - Vertex at bottom right
K - Vertex at top left of blue quadrilateral
C - vertex at top left of brown quadrilateral
L - vertex at top right of blue quadrilateral
F - vertex at top right of brown quadrilateral.
Line Segment AB is fixed.
Suppose that $$|AC| = |CF| = |FB|$$ and quadrilateral ACFB is convex.
Construct the unique quadrilateral AKLB such that $$ |AKLB| = |ACFB|, \quad \ |AK| = |KL| = |LB| \quad \text{and}\ \measuredangle{AKL} = \measuredangle KLB.$$
Given any function $f \in W^{1,2}(ACFB)$, consider the Rayleigh quotient: $$ \frac{\int_{ACFB} |\nabla f|^2 \ dx }{\int_{ACFB} |f|^2 \ dx}$$
Question: Is there a nice (and invertible) change of variable from $T:ACFB \mapsto AKLB$ such that under this transformation, the new function $f^*(y) := f(T^{-1}y)$ for $y \in AKLB$ satisfies $$ \frac{\int_{AKLB} |\nabla f^*|^2 \ dx }{\int_{AKLB} |f^*|^2 \ dx} \leq \frac{\int_{ACFB} |\nabla f|^2 \ dx }{\int_{ACFB} |f|^2 \ dx}$$
Notation: $|AB|$ denotes length of the side $AB$ whereas $|ACFB|$ denotes the area of quadrilateral $ACFB$.
In particular, can I use the Homography transformation to get this inequality?
