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No threshold can work uniformly over all holomorphic vector bundles $E$. Given a proposed threshold $M$, choose an integer $m\ge M$ and let $E=F^m$. The canonical trivialization of a line bundle tensored with its dual givesThis contradicts vanishing at that value of $m$. In the proof above, the curvature of $E=F^m$ exactly offsets the curvature contribution from $F^{-m}$, explaining why the bound cannot be independent of $E$.

Fibrewise evaluation identifiesThe identity endomorphism is a nowhere-zero continuous section, giving the canonical trivialization of a line bundle tensored with its dual.

Give a complex line bundle its natural orientation as a real plane bundle. Then its Euler class equals its First Chern class. If $a,b$ are the standard generators of $H^2(S^2\times S^2;\mathbb{Z})$, choose complex line bundles $L_1,L_2$ pulled back from the Hopf fibration on the two factors, with $e(L_1)=a$ and $e(L_2)=b$. The triviality of $L_i^{\ast }\otimes L_i$ gives $e(L_i^{\ast })=-e(L_i)$. Hence, for anythe tensor product $L_1^{\otimes p}\otimes L_2^{\otimes q}$, with negative powers interpreted using dual bundles, has Euler class $\alpha$. Its underlying oriented real rank-two bundle is the required $E_{\alpha }$.