= Solution
The <rank-nullity theorem> says that for a linear map $T:V\to W$ with finite-dimensional domain, $\dim V=\dim\ker T+\dim\operatorname{im}T$. Apply it to $\beta|_{\operatorname{im}\alpha}$. Its image is $\operatorname{im}(\beta\alpha)$ and its kernel is $\operatorname{im}\alpha\cap\ker\beta$, so
$$
\boxed{\dim\operatorname{im}\alpha
=\dim\operatorname{im}(\beta\alpha)
+\dim(\operatorname{im}\alpha\cap\ker\beta).}
$$
Since $\operatorname{im}(\alpha\gamma)\subseteq\operatorname{im}\alpha$,
$$
\dim(\operatorname{im}(\alpha\gamma)\cap\ker\beta)
\leq\dim(\operatorname{im}\alpha\cap\ker\beta).
$$
Using the preceding identity for both restrictions and rearranging gives
$$
\boxed{\dim\operatorname{im}(\beta\alpha)+\dim\operatorname{im}(\alpha\gamma)
\leq\dim\operatorname{im}\alpha+\dim\operatorname{im}(\beta\alpha\gamma).}
$$
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