[proofplan] Choose an arbitrary Hermitian metric on $F$ and a positive Hermitian metric on $L$. The curvature of $F\otimes L^m$ is the fixed curvature of $F$ plus $m$ times the positive line-bundle curvature acting as the identity on $F$. Compactness turns the fixed term into a uniform lower bound and the positive line-bundle term into a uniform positive lower bound. For $m$ sufficiently large, the positive term dominates the fixed negative part, giving Nakano positivity; the cohomology vanishing then follows from the [harmonic representative vanishing criterion](/theorems/9103). [/proofplan]

[step:Choose metrics and record the curvature formula] Let $r=\operatorname{rank}F$. Choose a Hermitian metric $h_F$ on $F$. Since $L$ is positive, choose a Hermitian metric $h_L$ on $L$ such that the real $(1,1)$-form \begin{align*} \omega_L:=i\Theta(L,h_L) \end{align*} is positive on $X$. For each integer $m\ge 1$, define the Hermitian metric \begin{align*} h_m:=h_F\otimes h_L^m \end{align*} on $F\otimes L^m$. By the [tensor product curvature formula](/theorems/9111) [citetheorem:9111], applied to $F$ and $L^m$ with their chosen metrics, \begin{align*} \Theta(F\otimes L^m,h_m)=\Theta(F,h_F)\otimes \operatorname{id}_{L^m}+\operatorname{id}_F\otimes \Theta(L^m,h_L^m). \end{align*} The curvature of the tensor power metric satisfies \begin{align*} \Theta(L^m,h_L^m)=m\Theta(L,h_L). \end{align*} Hence, after identifying the line factor with scalar multiplication on the fibre of $F\otimes L^m$, \begin{align*} i\Theta(F\otimes L^m,h_m)=i\Theta(F,h_F)+m\,\omega_L\otimes \operatorname{id}_F. \end{align*} [/step]

[step:Extract uniform curvature bounds from compactness]We measure tensors using the Kähler metric $\omega$ on $T^{1,0}X$ and the metric $h_F$ on $F$. Define the unit sphere bundle \begin{align*} S:=\{(x,u)\mid x\in X,\ u\in T_x^{1,0}X\otimes F_x,\ |u|_{\omega,h_F}=1\}. \end{align*} For $x\in X$ and $u=\sum_{j,\alpha}u_{j\alpha}\,\partial_{z_j}\otimes e_\alpha\in T_x^{1,0}X\otimes F_x$, written in a unitary holomorphic coordinate frame for $\omega$ and a unitary frame for $h_F$, define \begin{align*} B_F(x,u):=\sum_{j,k,\alpha,\beta} i\Theta(F,h_F)_{j\bar k\alpha\bar\beta}(x)\,u_{j\alpha}\,\overline{u_{k\beta}}. \end{align*} This number is independent of the chosen unitary frames because it is the Hermitian quadratic form on $T_x^{1,0}X\otimes F_x$ induced by the curvature tensor. The function $B_F:S\to \mathbb R$ is continuous, and $S$ is compact because $X$ is compact and $S$ is the unit sphere bundle of a finite-rank Hermitian vector bundle over $X$. Therefore $B_F$ has a minimum. Choose \begin{align*} A:=\max\{0,-\min_S B_F\}. \end{align*} Then, for every $x\in X$ and every $u\in T_x^{1,0}X\otimes F_x$, \begin{align*} B_F(x,u)\ge -A|u|_{\omega,h_F}^2. \end{align*} Similarly, positivity of $\omega_L$ means that the Hermitian form $\omega_L$ on $T_x^{1,0}X$ is positive definite at every $x\in X$. Define \begin{align*} S_X:=\{(x,\xi)\mid x\in X,\ \xi\in T_x^{1,0}X,\ |\xi|_\omega=1\}. \end{align*} The [continuous function](/page/Continuous%20Function) $(x,\xi)\mapsto \omega_L(\xi,\bar\xi)$ on the [compact space](/page/Compact%20Space) $S_X$ has a positive minimum. Define \begin{align*} a:=\min_{S_X}\omega_L(\xi,\bar\xi)>0. \end{align*} Then, for every $x\in X$ and every $\xi\in T_x^{1,0}X$, \begin{align*} \omega_L(\xi,\bar\xi)\ge a|\xi|_\omega^2. \end{align*}[/step]

[guided]The point of this step is to turn pointwise curvature information into constants that work everywhere on $X$. The fixed bundle $F$ may have curvature with negative directions, but compactness ensures that this negativity is uniformly bounded. We first use the Kähler metric $\omega$ and the Hermitian metric $h_F$ to measure tensors in $T^{1,0}X\otimes F$. Define \begin{align*} S:=\{(x,u)\mid x\in X,\ u\in T_x^{1,0}X\otimes F_x,\ |u|_{\omega,h_F}=1\}. \end{align*} For a point $x\in X$ and a tensor \begin{align*} u=\sum_{j,\alpha}u_{j\alpha}\,\partial_{z_j}\otimes e_\alpha\in T_x^{1,0}X\otimes F_x, \end{align*} written in a unitary frame, define \begin{align*} B_F(x,u):=\sum_{j,k,\alpha,\beta} i\Theta(F,h_F)_{j\bar k\alpha\bar\beta}(x)\,u_{j\alpha}\,\overline{u_{k\beta}}. \end{align*} Although this formula uses a frame, the value does not depend on the chosen unitary frame: it is precisely the curvature Hermitian form evaluated on the tensor $u$. The function $B_F:S\to\mathbb R$ is continuous because the Chern curvature tensor of the smooth Hermitian metric $h_F$ has smooth coefficients. The space $S$ is compact because it is the unit sphere bundle of the finite-rank Hermitian vector bundle $T^{1,0}X\otimes F$ over the compact base $X$. Therefore $B_F$ attains a minimum. Define \begin{align*} A:=\max\{0,-\min_S B_F\}. \end{align*} This choice gives $B_F(x,u)\ge -A$ whenever $|u|_{\omega,h_F}=1$. For an arbitrary nonzero $u$, apply this inequality to $u/|u|_{\omega,h_F}$ and multiply by $|u|_{\omega,h_F}^2$; for $u=0$ the inequality is immediate. Thus, for every $x\in X$ and every $u\in T_x^{1,0}X\otimes F_x$, \begin{align*} B_F(x,u)\ge -A|u|_{\omega,h_F}^2. \end{align*} Now we do the same for the positive curvature of $L$. Define \begin{align*} S_X:=\{(x,\xi)\mid x\in X,\ \xi\in T_x^{1,0}X,\ |\xi|_\omega=1\}. \end{align*} Since $\omega_L=i\Theta(L,h_L)$ is positive, $\omega_L(\xi,\bar\xi)>0$ for every nonzero $\xi\in T_x^{1,0}X$. The function $(x,\xi)\mapsto \omega_L(\xi,\bar\xi)$ is continuous on the compact set $S_X$, so it has a minimum. That minimum is strictly positive because the function is positive at every point of $S_X$. Set \begin{align*} a:=\min_{S_X}\omega_L(\xi,\bar\xi)>0. \end{align*} Scaling again gives, for every $x\in X$ and every $\xi\in T_x^{1,0}X$, \begin{align*} \omega_L(\xi,\bar\xi)\ge a|\xi|_\omega^2. \end{align*}[/guided]

[step:Choose the tensor power so the positive line curvature dominates] Let \begin{align*} m_0:=\left\lfloor \frac{A}{a}\right\rfloor+1. \end{align*} Fix an integer $m\ge m_0$. Let $x\in X$ and let \begin{align*} u=\sum_{j,\alpha}u_{j\alpha}\,\partial_{z_j}\otimes e_\alpha\in T_x^{1,0}X\otimes F_x \end{align*} be written in unitary frames for $\omega$ and $h_F$. The line-bundle contribution satisfies \begin{align*} \sum_{j,k,\alpha} m\,\omega_{L,j\bar k}(x)\,u_{j\alpha}\,\overline{u_{k\alpha}}\ge ma\sum_{j,\alpha}|u_{j\alpha}|^2=ma|u|_{\omega,h_F}^2. \end{align*} Combining this estimate with the lower bound for $B_F$ gives \begin{align*} \sum_{j,k,\alpha,\beta} i\Theta(F\otimes L^m,h_m)_{j\bar k\alpha\bar\beta}(x)\,u_{j\alpha}\,\overline{u_{k\beta}} \ge (ma-A)|u|_{\omega,h_F}^2. \end{align*} Since $m\ge m_0$, we have $ma-A>0$. Therefore $i\Theta(F\otimes L^m,h_m)$ is Nakano positive at every point of $X$. This proves the first assertion. [/step]

[step:Apply Nakano positivity to harmonic representatives] For $m\ge m_0$, set \begin{align*} E_m:=F\otimes L^m \end{align*} with the Nakano positive metric $h_m$. Since $X$ is compact Kähler and $E_m$ is a Hermitian holomorphic vector bundle, the Nakano lower bound from the previous step gives a constant \begin{align*} c_m:=ma-A>0 \end{align*} such that the curvature commutator satisfies the coercive estimate required in [citetheorem:9101] for $E_m$-valued $(n,q)$-forms, for every $q\ge 1$. Hence [citetheorem:9103] applies and every harmonic $E_m$-valued $(n,q)$-form is zero. By the Hodge identification for holomorphic vector bundles [citetheorem:9104], \begin{align*} H^q(X,K_X\otimes E_m)=0. \end{align*} Substituting $E_m=F\otimes L^m$ gives \begin{align*} H^q(X,K_X\otimes F\otimes L^m)=0 \end{align*} for every $q\ge 1$ and every $m\ge m_0$. [/step]