theorem
  a<b implies a/b<(b+sqrt((a^2+b^2)/2))/(a+sqrt((a^2+b^2)/2))
proof
  assume
A1: a<b;
  then
A2: sqrt(a/b)<(b+sqrt((a^2+b^2)/2))/(a+sqrt((a^2+b^2)/2)) by Th5;
  a/b<sqrt(a/b) by A1,Th4;
  hence thesis by A2,XXREAL_0:2;
end;
