reserve a,b,c,x,y,z for Real;

theorem
 a >= 0 implies sqrt(a+b^2) >= b
proof
 assume
A1: a >= 0;
 per cases;
 suppose b < 0;
  hence thesis by A1,Def2;
 end;
 suppose
A2: b >= 0;
A3: b^2 >= 0 by XREAL_1:63;
  a+b^2 >= 0+b^2 by A1,XREAL_1:6;
  then sqrt(a+b^2) >= sqrt(b^2) by A3,Th26;
  hence sqrt(a+b^2) >= b by A2,Def2;
 end;
end;
