Lemma 12.25.135DAA9B186CEE

Proof. It suffices to treat one nonzero holomorphic section $s$, since nonnegative real combinations preserve the inequalities. Choose a point at which $s$ is nonzero and which lies outside all exceptional divisors, and fix the coordinates of the other factors. A projective line through that point inside a fiber of $q_ i$ has $H_ i$-degree one and $Q_ i$-degree zero. The restriction of $s$ to it is a nonzero section, so the exponent of $\mathcal O(H_ i)$ is nonnegative. On a plain factor, a line through the point gives the same inequality.

On a blown-up factor, write the chosen point as $([z:w],[w])$ with $w\ne 0$, and choose a linear map $A$ with $Aw=z$. The section of $q_ i$ given by $[t]\mapsto ([At:t],[t])$ passes through the point, and both $H_ i$ and $Q_ i$ have degree one on each projective line in its image. Such a line through the point exists because $r_ i\geq 1$, and $s$ restricts nontrivially to it. Its degree shows that the sum of the two exponents is nonnegative.

Conversely, let $F_ i$ be the pullback under $q_ i$ of a hyperplane. An admissible pair $(\delta _ i,\rho _ i)$ is realized by $\delta _ iE_ i+(\delta _ i+\rho _ i)F_ i$, since $E_ i$ has curvature $H_ i-Q_ i$, and coefficients on plain factors are realized by hyperplanes. These are allowed nonnegative combinations and give every coefficient tuple in the stated range. $\square$