If $L$ is positive, $E\otimes L^k$ is Nakano positive for some $k$ I'm trying to prove the following:

Let $L$ be a positive holomorphic line bundle on a compact complex manifold $X$. For any hermitian holomorphic vector bundle $E$ on $X$, there is $k \in \mathbb{N}$ such that $E\otimes L^k$ is Nakano positive.

I'm using the decomposition given by diverietti in this answer. Part of my problem is that (I think) I can find a sufficiently large choice of $k$ so that $E\otimes L^k$ is Nakano positive at a point, but I can't extend beyond that. If I try to prove that I can find such a $k$ locally, my bounds are no longer valid. Maybe I need to use the fact that $X$ is locally compact.

Does anyone have a reference (or proof) of the above fact?


Update: I'm not actually sure that the statement I'm trying to prove is true, though I feel like it should be.
 A: This follows by the following general, quite elementary  
Fact. Suppose you have a complex vector bundle $E\to X$ on a compact manifold $X$ and two hermitian form $h_1$ and $h_2$ on $E$, such that the first one is positive definite. Then, there exists a constant $C_0>0$ such that for all $C\ge C_0$ the hermitian form $C\,h_1+h_2$ is positive definite.
Proof. Since both $h_1$ and $h_2$ are homogeneous, it suffices to check the statement on the unitary (with respect to $h_1$) bundle $U(E,h_1)\to X$, defined as the set of elements in $E$ of $h_1$-norm equal to one. $U(E,h_1)$ is obviously compact, since $X$ is. Let $m_i$ be the minimum of $h_i$ on $U(E,h_1)$. By compactness, $m_1>0$. We can also suppose $m_2<0$, otherwise we are done by taking any $C_0>0$. Now, take $C_0$ to be any real number $>-m_2/m_1$. Then, for any $v\in E\setminus\{0\}$, you have
$$
\begin{aligned}
C_0\,h_1(v,v)+h_2(v,v) & =||v||^2_{h_1}\bigl(C_0\,h_1(v/||v||_{h_1},v/||v||_{h_1})+h_2(v/||v||_{h_1},v/||v||_{h_1})\bigr) \\
& \ge ||v||^2_{h_1}(C_0\,m_1+m_2)>0. 
\end{aligned}
$$      
Now, remark that $h_E\otimes\theta_{L}$ defines a positive definite hermitian form on $T_X\otimes E\otimes L$ (this will be morally our $h_1$) since by hypothesis $(L,h_L)$ is positive, then apply the above fact with the other hermitian form defined by $\theta_E\otimes h_L$ (this will be morally our $h_2$). Let $C_0$ be the constant you get such that
$$
C_0\,h_E\otimes\theta_{L}+\theta_E\otimes h_L
$$
is positive definite on $T_X\otimes E\otimes L$ and $k$ any integer greater than $C_0$. 
Next, by the decomposition you cited, we have
$$
\theta_{E\otimes L^{\otimes k}}=\theta_E\otimes h_{L^{\otimes k}}+h_E\otimes\theta_{L^{\otimes k}}
$$ 
Claim. The above expression defines a positive hermitian form on $T_X\otimes E\otimes L^{\otimes k}$. Consequently $E\otimes L^{\otimes k}$ is Nakano positive. 
Proof. Without loss of generality, we can suppose $k\ge 2$. Then, the claim follows from the following elementary identity (which will be proven subsequently):
$$
\theta_{L^{\otimes k}}=k\,\theta_L\otimes h_{L^{\otimes(k-1)}}.
$$
Indeed, you now get
$$
\begin{aligned}
\theta_{E\otimes L^{\otimes k}} &=
\theta_E\otimes h_{L^{\otimes k}}+h_E\otimes\theta_{L^{\otimes k}}\\
&= \theta_E\otimes h_L\otimes h_{L^{\otimes(k-1)}}+k\,h_E\otimes \theta_L\otimes h_{L^{\otimes(k-1)}}\\
&=(\theta_E\otimes h_L+k\,h_E\otimes \theta_L)\otimes h_{L^{\otimes(k-1)}}\\
&\ge(\theta_E\otimes h_L+C_0\,h_E\otimes\theta_{L})\otimes h_{L^{\otimes(k-1)}},
\end{aligned}
$$
and the latter is clearly a positive definite hermitian form on $T_X\otimes E\otimes L^{\otimes k}$.
Coming back to the identity, notice first that whenever $F$ is a line bundle, on $T_X\otimes F$ we have
$$
\theta_F=h_F(\Theta_F(\bullet,\overline\bullet)\,\bullet,\bullet)=\Theta_F\otimes h_F,
$$
since $\Theta_F$ is just a $(1,1)$-form (seen as a hermitian operator on $T_X$).
Thus, we obtain
$$
\begin{aligned}
\theta_{L^{\otimes k}} &=\Theta_{L^{\otimes k}}\otimes h_{L^{\otimes k}}\\
& =k\,\Theta_{L}\otimes h_L\otimes h_{L^{\otimes(k-1)}}\\
&=k\,\theta_L\otimes h_{L^{\otimes(k-1)}}.
\end{aligned}
$$
