There is another inequality which says the following:

**Easy addition**
(Using the same notation):
$$
\kappa(X)\leqslant \kappa(X_y) + \dim Y
$$

Consequently, if $Y$ is of general type, i.e., $\kappa(Y)=\dim Y$, then the subadditivity conjecture is equivalent with equality instead of inequality.

Subadditivity is also known in the case $X_y$ is of general type (due to Kollár) and of course the inequality is trivially an equality (without easy addition) if both $X_y$ and $Y$ are of general type.

A problematic case is when $\kappa(Y)<0$. Then subadditivity doesn't say much and in general one cannot expect equality (although various hyperbolicity results can help). For instance, any elliptic K3 surface is fibred over $\mathbb P^1$ and gives an example when the inequality is strict.

One should probably add that there is a more sophisticated version of subadditivity (due to Viehweg) which predicts that if $\kappa(Y)\geq 0$, then
$$
\kappa(X)\geqslant\kappa(X_y) + \mathrm{max}\{\kappa(Y), \mathrm{Var}(f)\},
$$
where $\mathrm{Var}(f)$ is the *variation* of $f$, i.e., the measure of how much the family varies in moduli (if there is a corresponding moduli space then this is the dimension of the image of the corresponding moduli map). So, if $\kappa(Y)\lneq \mathrm{Var}(f)$, then the inequality in the original format is expected to be strict.