= Solution
If the first trial is accepted, $\alpha_k=\alpha_{\mathrm{init}}$. Otherwise the preceding trial was rejected and so exceeded $\alpha_{\max}$, since part (ii) guarantees acceptance below that threshold. The accepted step, reduced by the factor $\tau$, consequently exceeds $\tau\alpha_{\max}$. In either case,
$$
\alpha_k\geq\min(\alpha_{\mathrm{init}},\tau\alpha_{\max}).
$$
The angle condition $-g_k^Tp_k\geq\varepsilon\|g_k\|_2\|p_k\|_2$ gives $\alpha_{\max}\geq2\varepsilon(1-\beta)\|g_k\|_2/(\gamma\|p_k\|_2)$. Hence the <backtracking line search> has the lower bound
$$
\boxed{\alpha_k\geq\min\left(\alpha_{\mathrm{init}},\frac{2\tau\varepsilon(1-\beta)\|g_k\|_2}{\gamma\|p_k\|_2}\right).}
$$
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