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

theorem
  0 <= a & 0 <= b implies sqrt(a+b) <= sqrt a + sqrt b
proof
  assume that
A1: 0 <= a and
A2: 0 <= b;
A3: 0 <= sqrt a by A1,Def2;
  0 <= sqrt(a*b) by A1,A2,Def2;
  then 0 <= sqrt a*sqrt b by A1,A2,Th29;
  then 0 <= 2*(sqrt a*sqrt b);
  then a + 0 <= a + 2*sqrt a*sqrt b by XREAL_1:6;
  then
A4: a + b <= a + 2*sqrt a*sqrt b + b by XREAL_1:6;
A5: 0 <= sqrt b by A2,Def2;
  sqrt(a + 2*sqrt a*sqrt b + b) = sqrt((sqrt a)^2 + 2*sqrt a*sqrt b + b)
  by A1,Def2
    .= sqrt((sqrt a)^2 + 2*sqrt a*sqrt b + (sqrt b)^2) by A2,Def2
    .= sqrt((sqrt a + sqrt b)^2)
    .= sqrt a + sqrt b by A3,A5,Def2;
  hence thesis by A1,A2,A4,Th26;
end;
