$\theta(x) :=\sum_{p\leq x}\log{p}$, if $\epsilon>0$, then how it is possible to write that- $$\sum_{x^{1-\epsilon}< p\leq x}\log{p} \geq (1-\epsilon) (\log{x}) (\pi(x)-\pi(x^{1-\epsilon}))?$$
Because, There is no prime between $\pi(x)$ and $\pi(x^{1-\epsilon})$, so, $(\pi(x)-\pi(x^{1-\epsilon}))=0$ when $\epsilon \leq 1$, so definitely I am missing something, can any one explain how can we get above inequality ?
The source of the question is below theorem-
