[proofplan] The proof has three parts. First, we identify the local weight of the Fubini-Study metric on the affine chart $U_0$ and apply the definition of the Chern curvature form of a Hermitian line bundle. Second, we compute the mixed complex Hessian of $\log(1+|z|^2)$ in the affine coordinates $z_1,\dots,z_n$. Finally, we test the resulting Hermitian matrix on an arbitrary nonzero vector $\xi \in \mathbb{C}^n$ and prove strict positivity by an elementary quadratic inequality. [/proofplan] [step:Apply the local curvature formula to the Fubini-Study weight] Let $e_0$ denote the standard holomorphic local frame of $\mathcal{O}_{\mathbb{P}^n}(1)$ over $U_0$. The Fubini-Study metric is represented in this frame by the weight \begin{align*} \varphi_0: U_0 &\to \mathbb{R} \\ [Z_0:\cdots:Z_n] &\mapsto \log(1+|z|^2), \end{align*} meaning that $|e_0|_{h_{\mathrm{FS}}}^2=\exp(-\varphi_0)$. By the definition of the first Chern form of a Hermitian holomorphic line bundle in a local holomorphic frame, \begin{align*} c_1(\mathcal{O}_{\mathbb{P}^n}(1),h_{\mathrm{FS}}) = \frac{i}{2\pi}\partial\bar\partial \varphi_0. \end{align*} Substituting $\varphi_0=\log(1+|z|^2)$ gives \begin{align*} c_1(\mathcal{O}_{\mathbb{P}^n}(1),h_{\mathrm{FS}}) = \frac{i}{2\pi}\partial\bar\partial \log(1+|z|^2). \end{align*} [/step] [step:Differentiate the logarithmic potential in affine coordinates] Define \begin{align*} S: \mathbb{C}^n &\to \mathbb{R} \\ z &\mapsto 1+|z|^2 =1+\sum_{\ell=1}^n z_\ell \bar z_\ell. \end{align*} Then $\varphi_0=\log S$ in the affine coordinates on $U_0$. For each $1 \leq j \leq n$, \begin{align*} \frac{\partial \varphi_0}{\partial z_j} = \frac{1}{S}\frac{\partial S}{\partial z_j} = \frac{\bar z_j}{S}. \end{align*} For $1 \leq j,k \leq n$, differentiating this expression with respect to $\bar z_k$ and using the quotient rule gives \begin{align*} \frac{\partial^2\varphi_0}{\partial z_j\,\partial\bar z_k} &= \frac{\partial}{\partial\bar z_k}\left(\frac{\bar z_j}{S}\right) \\ &= \frac{\delta_{jk}S-\bar z_j z_k}{S^2} \\ &= \frac{(1+|z|^2)\delta_{jk}-\bar z_j z_k}{(1+|z|^2)^2}. \end{align*} Since \begin{align*} \partial\bar\partial \varphi_0 = \sum_{j,k=1}^n \frac{\partial^2\varphi_0}{\partial z_j\,\partial\bar z_k} \, dz_j \wedge d\bar z_k, \end{align*} we obtain \begin{align*} \partial\bar\partial \log(1+|z|^2) = \sum_{j,k=1}^n \frac{(1+|z|^2)\delta_{jk}-\bar z_j z_k}{(1+|z|^2)^2} \, dz_j \wedge d\bar z_k. \end{align*} [guided] The only computation needed is the complex Hessian of the local potential. We separate out the denominator by defining \begin{align*} S: \mathbb{C}^n &\to \mathbb{R} \\ z &\mapsto 1+|z|^2 =1+\sum_{\ell=1}^n z_\ell \bar z_\ell. \end{align*} Thus $\varphi_0=\log S$. The complex variables $z_j$ and $\bar z_j$ are treated as independent variables under Wirtinger differentiation, so \begin{align*} \frac{\partial S}{\partial z_j}=\bar z_j, \qquad \frac{\partial S}{\partial\bar z_k}=z_k. \end{align*} Therefore \begin{align*} \frac{\partial \varphi_0}{\partial z_j} = \frac{1}{S}\frac{\partial S}{\partial z_j} = \frac{\bar z_j}{S}. \end{align*} Now differentiate once more: \begin{align*} \frac{\partial^2\varphi_0}{\partial z_j\,\partial\bar z_k} &= \frac{\partial}{\partial\bar z_k}\left(\frac{\bar z_j}{S}\right). \end{align*} The numerator derivative is $\partial\bar z_j/\partial\bar z_k=\delta_{jk}$, and the denominator derivative is $\partial S/\partial\bar z_k=z_k$. Hence the quotient rule gives \begin{align*} \frac{\partial^2\varphi_0}{\partial z_j\,\partial\bar z_k} &= \frac{\delta_{jk}S-\bar z_j z_k}{S^2} \\ &= \frac{(1+|z|^2)\delta_{jk}-\bar z_j z_k}{(1+|z|^2)^2}. \end{align*} Finally, by the coordinate expression for $\partial\bar\partial$ on a smooth scalar function, \begin{align*} \partial\bar\partial \varphi_0 = \sum_{j,k=1}^n \frac{\partial^2\varphi_0}{\partial z_j\,\partial\bar z_k} \, dz_j \wedge d\bar z_k. \end{align*} Substituting the mixed derivative gives the displayed coefficient matrix. [/guided] [/step] [step:Test the coefficient matrix on an arbitrary tangent vector] Fix $z \in \mathbb{C}^n$. Define the Hermitian matrix \begin{align*} A(z) := \left( \frac{(1+|z|^2)\delta_{jk}-\bar z_j z_k}{(1+|z|^2)^2} \right)_{j,k=1}^n. \end{align*} Let $\xi=(\xi_1,\dots,\xi_n) \in \mathbb{C}^n$. Then \begin{align*} \sum_{j,k=1}^n A(z)_{jk}\xi_j\bar\xi_k &= \frac{1}{(1+|z|^2)^2} \sum_{j,k=1}^n \left((1+|z|^2)\delta_{jk}-\bar z_j z_k\right) \xi_j\bar\xi_k \\ &= \frac{(1+|z|^2)\sum_{j=1}^n|\xi_j|^2 - \left(\sum_{j=1}^n \bar z_j\xi_j\right) \left(\sum_{k=1}^n z_k\bar\xi_k\right)} {(1+|z|^2)^2} \\ &= \frac{(1+|z|^2)|\xi|^2-\left|\sum_{j=1}^n \bar z_j\xi_j\right|^2} {(1+|z|^2)^2}. \end{align*} The scalar $\sum_{j=1}^n \bar z_j\xi_j$ satisfies the elementary estimate \begin{align*} \left|\sum_{j=1}^n \bar z_j\xi_j\right|^2 \leq \left(\sum_{j=1}^n |z_j|^2\right) \left(\sum_{j=1}^n |\xi_j|^2\right) = |z|^2|\xi|^2. \end{align*} Therefore \begin{align*} \sum_{j,k=1}^n A(z)_{jk}\xi_j\bar\xi_k &\geq \frac{(1+|z|^2)|\xi|^2-|z|^2|\xi|^2} {(1+|z|^2)^2} \\ &= \frac{|\xi|^2}{(1+|z|^2)^2}. \end{align*} If $\xi \neq 0$, then $|\xi|^2>0$, so \begin{align*} \sum_{j,k=1}^n A(z)_{jk}\xi_j\bar\xi_k>0. \end{align*} Thus $A(z)$ is positive definite for every $z \in \mathbb{C}^n$. [guided] To prove positivity of the $(1,1)$-form, we test its coefficient matrix against an arbitrary vector. Fix $z \in \mathbb{C}^n$, and define \begin{align*} A(z) := \left( \frac{(1+|z|^2)\delta_{jk}-\bar z_j z_k}{(1+|z|^2)^2} \right)_{j,k=1}^n. \end{align*} Let $\xi=(\xi_1,\dots,\xi_n) \in \mathbb{C}^n$. The associated Hermitian quadratic form is \begin{align*} \sum_{j,k=1}^n A(z)_{jk}\xi_j\bar\xi_k. \end{align*} Substituting the entries of $A(z)$ gives \begin{align*} \sum_{j,k=1}^n A(z)_{jk}\xi_j\bar\xi_k &= \frac{1}{(1+|z|^2)^2} \sum_{j,k=1}^n \left((1+|z|^2)\delta_{jk}-\bar z_j z_k\right) \xi_j\bar\xi_k. \end{align*} The Kronecker delta term collapses the double sum to the norm square of $\xi$: \begin{align*} \sum_{j,k=1}^n \delta_{jk}\xi_j\bar\xi_k = \sum_{j=1}^n |\xi_j|^2 = |\xi|^2. \end{align*} The rank-one term factors as \begin{align*} \sum_{j,k=1}^n \bar z_j z_k\xi_j\bar\xi_k = \left(\sum_{j=1}^n \bar z_j\xi_j\right) \left(\sum_{k=1}^n z_k\bar\xi_k\right) = \left|\sum_{j=1}^n \bar z_j\xi_j\right|^2. \end{align*} Therefore \begin{align*} \sum_{j,k=1}^n A(z)_{jk}\xi_j\bar\xi_k = \frac{(1+|z|^2)|\xi|^2-\left|\sum_{j=1}^n \bar z_j\xi_j\right|^2} {(1+|z|^2)^2}. \end{align*} The only possible obstruction to positivity is the negative rank-one term. It is controlled by the elementary inequality \begin{align*} \left|\sum_{j=1}^n \bar z_j\xi_j\right|^2 \leq \left(\sum_{j=1}^n |z_j|^2\right) \left(\sum_{j=1}^n |\xi_j|^2\right) = |z|^2|\xi|^2. \end{align*} Substituting this estimate into the numerator gives \begin{align*} (1+|z|^2)|\xi|^2-\left|\sum_{j=1}^n \bar z_j\xi_j\right|^2 \geq (1+|z|^2)|\xi|^2-|z|^2|\xi|^2 = |\xi|^2. \end{align*} Hence \begin{align*} \sum_{j,k=1}^n A(z)_{jk}\xi_j\bar\xi_k \geq \frac{|\xi|^2}{(1+|z|^2)^2}. \end{align*} If $\xi \neq 0$, then $|\xi|^2>0$, and the denominator $(1+|z|^2)^2$ is also positive. Thus the quadratic form is strictly positive on every nonzero $\xi$, which is exactly positive definiteness of $A(z)$. [/guided] [/step] [step:Conclude positivity of the Fubini-Study Chern form] The previous step proves that the coefficient matrix of $\partial\bar\partial\log(1+|z|^2)$ is positive definite at every point of $U_0$. Since \begin{align*} c_1(\mathcal{O}_{\mathbb{P}^n}(1),h_{\mathrm{FS}}) = \frac{i}{2\pi}\partial\bar\partial\log(1+|z|^2), \end{align*} the first Chern form is a positive real $(1,1)$-form on $U_0$. The same computation holds on each standard affine chart by replacing the distinguished homogeneous coordinate $Z_0$ with the nonzero coordinate defining that chart. Therefore the Fubini-Study Chern form is positive on all of $\mathbb{P}^n$, and $\mathcal{O}_{\mathbb{P}^n}(1)$ equipped with $h_{\mathrm{FS}}$ is a positive Hermitian line bundle. [/step]