Given a Kähler manifold $X$, the blow up is a Kähler manifold aswell (see Hodge theory and CAG by Voision Prop. 3.24). The idea is of course to use the pull-back the Kähler form $\pi^*\omega_X$. It says that this form is not positive, but only semi-positive and I fail to see why exactly.
Positivity of $\omega_x$ means that locally we can write $\omega_x= \sum_i \alpha_i dz_i\wedge d\bar z_i$ with $\alpha_i$ real and non-negative. Now $\pi^*\omega_X= \sum_i (\alpha_i\circ\pi) d(z_i\circ\pi)\wedge d(\bar z_i\circ\pi)$. Obviously the coefficients $(\alpha_i\circ\pi)$ still remain positive, so somehow the differential forms $ d(z_i\circ\pi)\wedge d(\bar z_i\circ\pi)$ must vanish on some special tangent vectors in a neighborhood around the blow up?
I am also not sure, if my notion of positivity is correct or appropriate here.
Let $\widetilde X$ be the blowup of $X$, and let $\pi\colon \widetilde X\to X$ denote the blow-down map. If the local coordinates $(z_i)$ are chosen so that $z=0$ is the point being blown up, then $D = \pi^{-1}(0)\subseteq \widetilde X$ is a complex hypersurface called the exceptional divisor. Because $z_i\circ\pi\equiv 0$, it follows that $d(z_i\circ \pi)$ annihilates every vector tangent to $D$, and the same goes for $d(\bar z_i\circ\pi)$.