theorem Th5:
  a<b implies sqrt(a/b)<(b+sqrt((a^2+b^2)/2))/(a+sqrt((a^2+b^2)/2))
proof
  assume
A1: a<b;
  then a/b<1 by XREAL_1:189;
  then
A2: sqrt(a/b)<1 by SQUARE_1:18,27;
  sqrt((a^2+b^2)/2)>0 by SQUARE_1:25;
  then 1<(b+sqrt((a^2+b^2)/2))/(a+sqrt((a^2+b^2)/2)) by A1,Th3;
  hence thesis by A2,XXREAL_0:2;
end;
