Question for the proof of $L^p$ space

This is simply $$ \frac{\|f\|_q^q}{\alpha^q}\geq\frac{\|f\|_p^p}{\alpha^p} \iff \alpha^{q-p}\leq \frac{\|f\|_q^q}{\|f\|_p^p} \iff \alpha\leq \left(\frac{\|f\|_q^q}{\|f\|_p^p}\right)^{\frac1{q-p}}=B. $$ Thus $$ \min\left\{\frac{\|f\|_q^q}{\alpha^q},\frac{\|f\|_p^p}{\alpha^p}\right\} =\begin{cases} \frac{\|f\|_p^p}{\alpha^p},&\ \alpha\leq B \\ \ \\ \frac{\|f\|_q^q}{\alpha^q},&\ \alpha>B \end{cases} $$

This is a classical technique in harmonic analysis with the title: let some constant to be determined later. You may try to put $B=1$ to see what happen, of course, you cannot conclude anything actually. So the strategy is to keep $B$ fixed and we will config the value of $B$ while needed.

If we want $\min\{\alpha^{-p}\|f\|_{p,\infty}^{p},\alpha^{-q}\|f\|_{q,\infty}^{q}\}=\alpha^{-p}\|f\|_{p,\infty}^{p}$, this means $\alpha^{-p}\|f\|_{p,\infty}^{p}\leq\alpha^{-q}\|f\|_{q,\infty}^{q}$, so $\dfrac{\|f\|_{q,\infty}^{q}}{\|f\|_{p,\infty}^{p}}\geq\dfrac{\alpha^{q}}{\alpha^{p}}=\alpha^{q-p}$, so $\alpha\leq B$, where we need the role of $B$ to be satisfying in this fashion.

Further note here: in elementary mathematical analysis, when we deal with integral like $\displaystyle\int_{0}^{\infty}\dfrac{e^{-x}}{x^{1/2}}dx$, one deal with random partition of the domain of the integral to be $[0,c]$ and $[c,\infty)$ and it still goes through: $\displaystyle\int_{0}^{c}\dfrac{e^{-x}}{x^{1/2}}dx+\int_{c}^{\infty}\dfrac{e^{-x}}{x^{1/2}}dx<\infty$. Unfortunately, when we deal with more delicate integral like your question, usually one cannot set $B$ to be arbitrary number, it depends heavily on the context.