I am reading the book "Elliptic Partial Differential Equations of Second Order" by D. Gilbarg and N.S. Trudinger.
In theorem 7.27, it stated that after we obtained $$|u'(x)| \leq \int_{a}^{b}|u''| + \frac{18}{\epsilon^2}\int|u|,$$ we can apply Holder's inequality to get $$|u'(x)|^p \leq 2^{p-1}\left\{ \epsilon^{p-1}\int_{a}^{b}|u''|^p + \frac{(18)^p}{\epsilon^{p+1}}\int_{a}^{b}|u|^p \right\}.$$
I am wondering how Holder's inequality is used to get the second inequality.
In that proof, $\epsilon=b-a$. So,
$$\begin{align} \int_{a}^{b}|u''|&=\int_{a}^{b}|1\cdot u''|\\ &\leq\left(\int_{a}^{b}|1|^q\right)^{1/q}\left(\int_{a}^{b}|u''|^p\right)^{1/p}\\ &=(b-a)^{1/q}\left(\int_{a}^{b}|u''|^p\right)^{1/p}\\ &=\epsilon^{(p-1)/p}\left(\int_{a}^{b}|u''|^p\right)^{1/p} \end{align}$$
Now, do a similar calculation for the integral
$$\frac{18}{\epsilon^2}\int|u|=\frac{1}{\epsilon^2}\int|1\cdot 18u|,$$
take the power $p$ and use the inequality $$(a+b)^p\leq 2^{p-1}(a^p+b^p).$$
