Interesting question
construction: let the midpoint of XY by P. let the radius be x units long.
Proof: To prove XY>AB and CD is the same as proving OP < OM and OL. as lines from origin to midpoints of chords are perpendicular to the chord and the closer the chord is to the origin the bigger it is.
now, XM.MY=AM.MB (intercepts on intersecting chords)
but M is midpoint of AB. therefore XM.MY=AM^2 .....(1)
similarly for other chord, XL.LY=CL^2 ....(2)
In triangle AMO, angleAMO=90 (midpoint perp to origin)
therefore, AM^2+OM^2=x^2 (pythagoras, radius is x units long)
and OM=sqrt(x^2-AM^2)=sqrt(x^2-XM.YM), from (1)
similarly for triangle OCL and from (2)
OL=sqrt(x^2-XL.LY)
similarly for triangle XOP,
OP=sqrt(x^2-XP^2)
we want to prove OP < OM and OL
from these equations we need to show that x^2-XP^2< x^2-XM.MY and x^2-XL.LY
or that XP^2>XM.MY and XL.LY
ie, that the product of two intersections of a segment is greatest when the intersection is taken from the midpoint. (not sure how to word it here, but hopefully its understandable what i mean.)
proof of this: let the line segment be E units long. take one section as y units, then the other will be (E-Y). let the product be F. therefore F=y(E-y)=Ey-y^2.
differentiating, F'=E-2y. F'=0 when y=E/2. ie product of the two sections is greatest when they are halved.
applying this to circle question, therefore XP^2 > XL.LY and XM.MY, which implies OP < OM and OL
and therefore XY > AB and CD
i know this is a pretty dodgey proof but i dont really want to spend too much time writing this. Also if anyone can think of a shorter proof, let me know.