TITLE: Relation between $h^1(\mathcal{O}_X)$, $h^0(\Omega_X)$, and first betti number for general complex manifold?
QUESTION [5 upvotes]: I was wondering in the world of non-Kähler compact complex manifold, what is the relationship between $h^1(\mathcal{O}_X)$, $h^0(\Omega_X)$, and first betti number?
The only thing I know is that for surfaces, we have $h^1(\mathcal{O}_X) \geq \frac{b_1}{2} \geq h^0(\Omega_X)$, the reason being that every holomorphic 1-form on a compact surface must be closed. Does this inequality hold true for general compact complex manifold? I'm mainly interested in the relation between $h^1(O_X)$ and $h^0(\Omega_X)$.
Another somewhat related question is: does there exist a compact complex manifold with non-closed holomorphic 1-forms?
Thanks to an anonymous friend's comment. On page 444 of Griffiths and Harris' book Principles of Algebraic Geometry, they gave an example (Iwasawa manifold) of a non-closed holomorphic 1-form.

REPLY [2 votes]: It feels awkward to answer the question posed by myself, but I was told it's okay to do so. There are few points I want to say here.
First of all, regarding Remy's question. On a compact complex surface $X$, every holomorphic 1-form is closed, hence $h^0(\Omega_X) \leq h^1(O_X)$ holds true in this case (by David Speyer's answer). So suppose $\eta$ is a holomorphic 1-form on $X$, then $\int_X d\eta \wedge \bar{d\eta}=0$ by Stokes. Then one expands $d\eta \wedge \bar{d\eta}$ in local coordinate to see that this would imply $d\eta=0$. More generally on a compact complex n-fold, every holomorphic (n-1)-form is closed by the same argument.
Secondly, my main question is actually whether the inequality $h^0(\Omega_X) \leq h^1(O_X)$ holds in general. Now I knew a counterexample to this inequality which is Iwasawa manifold. One can just consult this and that. Probably I should describe Iwasawa manifold here in case someone forgot the definition. Let 
$G=\{\begin{bmatrix}
    1       &a          & b  \\
    0       &1          & c  \\
    0       &0          & 0
\end{bmatrix}|a, b, c \in \mathbb{C}\}$ 
be the Heisenberg group of complex coefficient and let $\Gamma$ be the subgroup with entries in $\mathbb{Z}[i]$, then we form a quotient $X=\Gamma \backslash G$. By projecting to $(a,c)$ coordinates, we exhibit $X$ as a holomorphic fibration with fiber $E$ and base $E \times E$, where $E$ is the beautiful elliptic with j-invariant 1728. One compute directly that $\Gamma^{ab}=\mathbb{Z}[i]^2$, hence $h^1(X,\mathbb{C})=4$. One sees (arguably not so obvious) that $da$, $dc$, and $db-adc$ form a basis of $H^0(X,\Omega^1_X)$. Applying Leray spectral sequence to the projection $X \to E \times E$, one can actually show that $h^1(O_X)=2$ (the point being that the differential $d_2: H^0(R^1\pi_*(O_X)=O_{E \times E}) \to H^2(O_{E\times E})$ is nonzero, and it sends 1 to $\bar{da} \wedge \bar{dc}$ if one normalizes everything appropriately).
Thirdly, Hopf surface would also give an example where the inequality $h^{01} \geq z^{10}$ in Davide Speyer's answer is not an equality ($z^{10}=h^{10}=0$ in that case).
Lastly, I was also wondering if there's an example where the Hodge-deRham doesn't degenerate at $E_2$ page. I was informed by my friend yesterday that such examples (nilmanifold) do exist, but I know nothing about it so perhaps I should stop here.