By the change of variables formula one gets

$$Vol_{\omega}(f^{j}K)=\int_{f^{j}K}\omega^{n}\geq\frac{1}{d_{t}^{j}}\int_{K}(f^{j})^{*}\omega^{n}=\frac{1}{d_{t}^{j}}\int_{K}|J_{FS}(f^{j})|^{2}\omega^{n},$$

where $d_{t}=\lambda^{n}$ denotes the topological degree of $f$ and $J_{FS}(f)$ stands for the jacobian of $f$ with respect to the Fubini-Study volume form. Observe that $\log|J_{FS}(f)|=u-v$ is a difference of two qpsh functions $u,v\in PSH(X,A\omega)$ for some $A=A(\lambda,n)$. Moreover by the chain rule,

$$\frac{1}{\lambda^{j}}\log|J_{FS}(f^{j})|=\sum_{l=0}^{j-1}\frac{1}{\lambda^{j}}\log|J_{FS}(f)\circ f^{l}|.$$

Since $\lambda^{-l}\log|J_{FS}(f)\circ f^{l}|$ is relatively compact in $L^{1}(\mathbb{C}\mathbb{P}^{n})$ (previous corollary), the concavity of the log yields

$$\frac{1}{Vol_{\omega}(K)}\int_{K}|J_{FS}(f^{j})|^{2}\omega^{n}\geq\exp\left(\frac{2\lambda^{j}}{Vol_{\omega}(K)}\int_{K}\frac{1}{\lambda^{j}}\log|J_{FS}(f^{j})|\omega^{n}\right)\geq\gamma^{\lambda^{j}}.$$

Decreasing slightly the value of $\gamma$ if necessary, this yields the desired inequality. ∎