We first prove

(6.1)

$$\max_{i\in I}|\varphi(e_{i})|\leq C.$$

Recall that we have normalized $\varphi$ so that $\max_{i\in I}\varphi(e_{i})=0$. We may therefore assume that $\mathcal{X}_{0}$ has at least two components. Note that $\pi_{*}G=\sum_{j}b_{j}\varphi(e_{j})E_{j}$ by definition. For each component $E_{i}$ the projection formula shows that

$$(\theta_{\mathcal{X}}+\pi_{*}G)\cdot E_{i}\cdot\mathcal{A}^{n-1}=(\pi^{*}\theta_{\mathcal{X}}+G)\cdot\pi^{*}E_{i}\cdot\pi^{*}\mathcal{A}^{n-1},$$

which is non-negative since $\pi^{*}\theta_{\mathcal{X}}+G$ and $\pi^{*}\mathcal{A}$ are nef and $\pi^{*}E_{i}\in\Div_{0}(\mathcal{Y})$ is effective. It follows that

(6.2)

$$\sum_{j}b_{j}\varphi(e_{j})(E_{i}\cdot E_{j}\cdot\mathcal{A}^{n-1})\geq-C$$

for all $i$. Note that $E_{i}\cdot E_{j}\cdot\mathcal{A}^{n-1}\geq 0$ for all $i\neq j$, and

$$b_{i}E_{i}\cdot E_{i}\cdot\mathcal{A}^{n-1}=E_{i}\cdot(b_{i}E_{i}-\mathcal{X}_{0})\cdot\mathcal{A}^{n-1}=-\sum_{j\neq i}b_{j}\,E_{i}\cdot E_{j}\cdot\mathcal{A}^{n-1}<0$$

for all $i$, since $\mathcal{X}_{0}$ has connected support and contains at least two components.

Now pick $i_{0},\dots,i_{M}$ such that $\varphi(e_{i_{0}})=0$, $\varphi(e_{i_{M}})=\min_{i\in I}\varphi(e_{i})$, and $e_{i_{m}}$ and $e_{i_{m+1}}$ are connected by a $1$-dimensional face, so that $E_{i_{m}}\cdot E_{i_{m+1}}\cdot\mathcal{A}^{n-1}\geq 1$. Writing

$$\lambda=\max_{i\in I}\left\{-b_{i}E_{i}^{2}\cdot\mathcal{A}^{n-1}\right\}>0$$

and applying (6.2) to $i_{0},i_{1},\dots,i_{M-1}$, we get by induction

$$0\geq\varphi(e_{i_{M}})\geq-C\sum_{m=1}^{M}\lambda^{m},$$

which proves (6.1).