As in the picture below, the rectangle on the left hand side has been divided into four parts by line segments which are parallel to a side of the rectangle. The areas of the small rectangles are A,B,C and D. Similarly, the small rectangles on the right hand side have areas A′,B′,C′ and D′. It is known that A≤A′, B≤B′, C≤C′ but D≤B′.
[asy]
import graph; size(12cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.3,xmax=12.32,ymin=-10.68,ymax=6.3;
draw((0,3)--(0,0)); draw((3,0)--(0,0)); draw((3,0)--(3,3)); draw((0,3)--(3,3)); draw((2,0)--(2,3)); draw((0,2)--(3,2)); label("A",(0.86,2.72),SE*lsf); label("B",(2.38,2.7),SE*lsf); label("C",(2.3,1.1),SE*lsf); label("D",(0.82,1.14),SE*lsf); draw((5,2)--(11,2)); draw((5,2)--(5,0)); draw((11,0)--(5,0)); draw((11,2)--(11,0)); draw((8,0)--(8,2)); draw((5,1)--(11,1)); label("A′",(6.28,1.8),SE*lsf); label("B′",(9.44,1.82),SE*lsf); label("C′",(9.4,0.8),SE*lsf); label("D′",(6.3,0.86),SE*lsf);
dot((0,3),linewidth(1pt)+ds); dot((0,0),linewidth(1pt)+ds); dot((3,0),linewidth(1pt)+ds); dot((3,3),linewidth(1pt)+ds); dot((2,0),linewidth(1pt)+ds); dot((2,3),linewidth(1pt)+ds); dot((0,2),linewidth(1pt)+ds); dot((3,2),linewidth(1pt)+ds); dot((5,0),linewidth(1pt)+ds); dot((5,2),linewidth(1pt)+ds); dot((11,0),linewidth(1pt)+ds); dot((11,2),linewidth(1pt)+ds); dot((8,0),linewidth(1pt)+ds); dot((8,2),linewidth(1pt)+ds); dot((5,1),linewidth(1pt)+ds); dot((11,1),linewidth(1pt)+ds);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]
Prove that the big rectangle on the left hand side has area smaller or equal to the area of the big rectangle on the right hand side, i.e. A+B+C+D≤A′+B′+C′+D′. inequalitiesgeometryrectanglecombinatorics unsolvedcombinatorics