# $\chi(\omega_X)>0$ implies that $X$ is of general type

If $X$ is a smooth (complex) projective variety of maximal Albanese dimension such that $\chi(\omega_X)>0$, how does one show that $X$ is of general type?

I've seen this used but I can't find a proof.

It is a result of Ein-Lazarsfeld that if $X$ is a smooth projective variety of maximal albanese dimension such that $\chi(\omega_X)=0$, then the albanese image is fibered by tori, so in partixular if $X$ is birational to its image under the albanese map, then it can't be of general type.

If it is not of general type, then the Iitaka fibration $X\to Z$ is fibered by tori say $F$ (the fibers of the Iitaka fibration have Kodaira dimension 0 and maximal Albanese dimension, and so by a Theorem of Kawamata, they are abelian varieties). For general $P\in Pic^0(A)$, we have that $P|_F\ne 0$ and so $h^0(K_F+P)=h^0(P|_F)=0$ where $F$ is the general fiber. But then $h^0(K_X+P)=0$ and by the generic vanishing theorems of Green and Lazarsfeld $h^i(K_X+P)=0$ for $i>0$ so that $\chi (K_X)=\chi(K_X+P)=0$.