Thanks Rohit, this is perfect. I was not aware that assumptions can use patterns.

BTW: The second parameter in Simplify is superfluous. Here some more examples how to use:

Block[{x, expr},
  expr = Sqrt[(1 + 2 Conjugate[x] + x^2)];
  Print[{expr, 1, Simplify[expr, Assumptions -> Element[_Symbol, Reals]]}];
  Print[{expr, 2, Simplify[expr, Element[_Symbol, Reals]]}];
  Print[{expr, 3, Assuming[Element[_Symbol, Reals], Simplify[expr]]}];
  $Assumptions = Element[_Symbol, Reals];
  Print[{expr, 4, Simplify[expr]}];
  ];

==>

{Sqrt[1+x^2+2 Conjugate[x]],1,Abs[1+x]}
{Sqrt[1+x^2+2 Conjugate[x]],2,Abs[1+x]}
{Sqrt[1+x^2+2 Conjugate[x]],3,Abs[1+x]}
{Sqrt[1+x^2+2 Conjugate[x]],4,Abs[1+x]}

Of course this works for many variables as well:

Block[{x, y, expr1, expr2},
  $Assumptions = True;
  expr1 = Sqrt[(1 + 2 Conjugate[x] + x^2)];
  expr2 = Sqrt[x^2 + 2 x Conjugate[y] + y^2];

  Print["Without assumptions: ", $Assumptions];
  Print[{expr1, Simplify[expr1]}];
  Print[{expr2, Simplify[expr2]}];

  Assuming[Element[_Symbol, Reals],
   Print["With assumptions: ", $Assumptions];
   Print[{expr1, Simplify[expr1]}];
   Print[{expr2, Simplify[expr2]}]
   ];
  ];

==>

Without assumptions: True
{Sqrt[1+x^2+2 Conjugate[x]],Sqrt[1+x^2+2 Conjugate[x]]}
{Sqrt[x^2+y^2+2 x Conjugate[y]],Sqrt[x^2+y^2+2 x Conjugate[y]]}
With assumptions: Element[_Symbol, Reals]
{Sqrt[1+x^2+2 Conjugate[x]],Abs[1+x]}
{Sqrt[x^2+y^2+2 x Conjugate[y]],Abs[x+y]}