reserve r1,r2,r3 for non negative Real;
reserve n,m1 for Nat;
reserve s for Real;
reserve cn,cd,i1,j1 for Integer;
reserve r for irrational Real;
reserve q for Rational;
reserve c0,c1,c2,u,a0,b0 for Real;
reserve a,b for Real;
reserve n for Integer;

theorem Th26:
   (n-a)*(n+1-b)>0 & (b-n)*(n+1-a)> 0 implies
   (n-a)*(n+1-b)+(b-n)*(n+1-a) = b-a &
   |.a-n.|*|.b-n.|*|.a-n-1.|*|.b-n-1.|<=|.a-b.|^2/4
   proof
     assume that
A1:  (n-a)*(n+1-b) > 0 and
A2:  (b-n)*(n+1-a) > 0;
     set s = (n-a)*(n+1-b);
     set t = (b-n)*(n+1-a);
A3:  sqrt(s*t) <= (s+t)/2 by A1,A2,SERIES_3:2;
A4:  (sqrt(s*t))^2 = s*t by A1, A2, SQUARE_1:def 2;
A5:  sqrt(s*t) >= 0 by A1, A2, SQUARE_1:def 2;
A6:  s=|.s.| by A1,COMPLEX1:43
     .=|.n-a.|*|.n+1-b.| by COMPLEX1:65;
A7:  t=|.t .| by A2,COMPLEX1:43
     .=|.b-n.|*|.n+1-a.| by COMPLEX1:65;
A8:  |.n-a .|=|. -(n-a).| by COMPLEX1:52 .= |. a-n.|;
A9:  |.n+1-b.|=|.-(n+1-b).| by COMPLEX1:52
     .= |.b -n-1.|;
A10: |.n+1-a.|=|.-(n+1-a).| by COMPLEX1:52
     .=|.a-n-1.|;
A11: s*t=|.n-a.|*|.n+1-b.|*(|.b-n.|*|.n+1-a.|) by A7,A6
     .=|.b-n.|*|.a-n .|*|.b-n-1.|*|.a-n-1.| by A10,A9,A8;
     (b-a)^2 = |.b-a.|^2 by COMPLEX1:75 .=|.a-b.|^2 by COMPLEX1:60; then
     ((b-a)/2)^2 =|.a-b.|^2/4;
     hence thesis by A3,A4,A5,A11,SQUARE_1:15;
   end;
