Just to get this off the unanswered list.
As mentioned in the original post, an effective divisor $D$ on $\mathbb{P}^n$ is a finite sum $\displaystyle \sum_i a_i V_i$ where $a_i\in\mathbb{Z}^{\geqslant 0}$ and $V_i$ are irreducible subvarieties of $\mathbb{P}^n$. We call such a divisor positive if the associated line bundle $\mathcal{O}(D)$ has a Hermitian metric with positive curvature. It's easy to see that if $D_1,\ldots,D_m$ are positive then so is $D_1+\cdots +D_m$.
So, to see that every effective divisor on $\mathbb{P}^n$ is positive it suffices to show that every divisor of the form $V$ is positive where $V$ is a closed analytic submanifold. But, by Chow's theorem we know that $V=V(f)$ where $f$ is a homogenous polynomial in $n+1$-variables of degree $d$. But, one can quickly check by hand that $V(f)$ is then equivalent to $dH$ where
$$\mathbb{P}^{n-1}\cong H:=\{[0:z_1:\cdots:z_n]\}\subseteq \mathbb{P}^n$$
So, by our above observations it suffices to show that $H$ is positive. But, this is clear.
One can also verify the claim that every $V$ is linearly equivalent to a multiple of $H$ without appealing to Chow's theorem as follows. We know that the line bundles on $\mathbb{P}^n$ are classified by $H^1(\mathbb{P}^n,\mathcal{O}^\times)$. But, by the exponential sequence
$$0\to \underline{2\pi i\mathbb{Z}}\to \mathcal{O}\to \mathcal{O}^\times\to 0$$
we get the exact sequence
$$H^1(\mathbb{P}^n,\underline{2\pi i\mathbb{Z}})\to H^1(\mathbb{P}^n,\mathcal{O})\to H^1(\mathbb{P}^n,\mathcal{O}^\times)\to H^2(\mathbb{P}^n,\underline{2\pi i \mathbb{Z}})\to H^2(\mathbb{P}^n,\mathcal{O})$$
But,
$$H^i(\mathbb{P}^n,\underline{2\pi i \mathbb{Z}})\cong H^i_\text{sing}(\mathbb{P}^n,\mathbb{Z})=\begin{cases} \mathbb{Z} & \mbox{if}\quad i\in\{0,2,\ldots,2n\}\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$$
and
$$H^i(\mathbb{P}^n,\mathcal{O})=\begin{cases} \mathbb{C} & \mbox{if}\quad i=0\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$$
from where we see that
$$\mathrm{Pic}(\mathbb{P}^n)\cong H^1(\mathbb{P}^n,\mathcal{O}^\times)\cong H^1(\mathbb{P}^n,\underline{2\pi i \mathbb{Z}})\cong \mathbb{Z}$$
call the map
$$H^1(\mathbb{P}^n,\mathcal{O}^\times)\to H^2(\mathbb{P}^n,\underline{2\pi i\mathbb{Z}})\cong \mathbb{Z}$$
the 'Chern class map' and denote it by $c$. What the above shows is that a line bundle on $\mathbb{P}^n$ is entirely determined by its Chern class.
Now, it's not hard to directly check that $c(\mathcal{O}(H))=1$ and in general $c(\mathcal{O}(D))>0$ if $D$ is effective (just think about the cycle class map) from which case it follows that $\mathcal{O}(D)\cong \mathcal{O}(H)^{c(\mathcal{O}(D))}$ or, in other words, $D\sim c(\mathcal{O}(D))H$ (where $\sim$ denotes equivalence).
Best Answer
No, any construction using the Hermitian metric is going to take you outside the holomorphic category! By the way, you can understand the Euler sequence very elegantly by mapping $\Bbb P^n\times \Bbb C^{n+1}\to T\Bbb P^n\otimes\mathscr L$ as follows: If $\pi\colon\Bbb C^{n+1}-\{0\}\to\Bbb P^n$ and $\pi(\tilde p) = p$, for $\xi\in T_{\tilde p}\Bbb C^{n+1}$, map $\xi$ to $\pi_{*\tilde p}\xi\otimes\tilde p$, and check this is well-defined.
You might also try to generalize the Euler sequence to a complex submanifold $M\subset\Bbb P^n$, letting $\tilde M = \pi^{-1}M$. Then you get the exact sequence $$0\to \mathscr L \to E \to TM\otimes\mathscr L\to 0\,,$$ where $E_p = T_{\tilde p}\tilde M$ for any $\tilde p\in \pi^{-1}(p)$. ($\tilde M$ is the affine cone corresponding to $M\subset\Bbb P^n$.)