Given a rank two tensor $T_{ab}$, we can symmetrize and antisymmetrize it as
$$T_{(ab)} = \frac{1}{2}(T_{ab}+T_{ba}), \quad T_{[ab]} = \frac{1}{2}(T_{ab}-T_{ba}).$$
In this case it is trivial to see that removing all parts of $T$ that are antisymmetric in any pair of indices, gives the totally symmetric part of $T$.
The same is true for higher rank tensors, as can be shown via Young tableaux/symmetrizers/etc. I am looking for the simplest way to show this, and am wondering whether the usual "technology" is actually necessary in this particular case.
What is the simplest way, or an elementary way, to see that a rank $r$ tensor can always be decomposed as a totally symmetric term, plus terms that are antisymmetric in the exchange of at least a pair of indices?