Are all three-point functions automatically zero? I'm switching between QFT books all the time. From Maggiore's chapter 5 I got the impression that 
$$ \langle \phi_1 \phi_2 \phi_3 \rangle \equiv \langle 0 |T\{\phi_1 \phi_2 \phi_3\}|0 \rangle =0 $$
because, on using Wick contraction to expand the time-ordered fields, there is always a spare un-contracted field which kills the term due to its action on the vacuum.
Now, following Schwartz I'm deriving the Schwinger-Dyson's equations in chapter 7. Here, he does not mention any such thing. In fact, he asks readers to calculate 
$$ \Box_x \langle\phi_1 \phi_2 \phi_x \rangle $$
Why isn't this object automatically zero?
 A: You have to be careful to compare the definitions between the books.
Maggiore is setting up perturbation theory in the usual way. To recap, the LSZ reduction theorem relates $S$-matrix elements to time-ordered correlators of Heisenberg fields. These may be related to time-ordered correlators of fields in the interaction picture by Maggiore's Eq. 5.67. And fields in the interaction picture always behave like free fields, so we may apply Wick's theorem to conclude that
$$\langle 0 | T \phi_I(x_1) \phi_I(x_2) \phi_I(x_3) | 0 \rangle = 0.$$
Since interaction picture perturbation theory is so common, the $I$ subscript is usually dropped, as Maggiore warns in the beginning of section 5.4.
Schwartz eventually does the same thing, in section 7.2, but he first derives the Feynman rules using a completely different route, by the Schwinger-Dyson equation in section 7.1. Here we simply work with Heisenberg fields the entire time, so 
$$\langle 0 | T \phi_H(x_1) \phi_H(x_2) \phi_H(x_3) | 0 \rangle$$
where the $H$ subscript is dropped. To add to this confusion, he refers to these fields as "the interacting fields", but only to emphasize they we're no longer doing free field theory, not that we're in interaction picture. Somewhat paradoxically the point of going to interaction picture is to make the interacting fields look free instead.
