Help needed in proving a result assuming Prime number theorem

Question

I need help in deducing a result.

Note that $0<\epsilon<1/3$ and $s$ be sufficiently large with respect to $\epsilon$.

Let $D$ be product of all primes less than or equal to $(1-2\epsilon)\log s$. Then it is to be proved assuming prime number theorem that $$\log(D) = \sum_{\substack{p \text{ prime}\\p\leq (1-2\epsilon) \log(s) }} \log p \leq (1-\epsilon) \log (s ).$$

My attempt: I tried by finding $\pi((1-\epsilon) \log (s))$ and then multiplied it by $(1-2(1-\epsilon) \log (s))$ to get an upper bound. But this is not what is asked in question. But what bounds I got are completely different from those asked in question.

Can someone please help.

Answer 1

By the prime number theroem, for any $0<\varepsilon<1/3$ there is an $x_0>0$ such that $$ \pi(x)\leq \frac{{1 - \varepsilon }}{{1 - 2\varepsilon }}\frac{x}{\log x} $$ (note that $\frac{{1 - \varepsilon }}{{1 - 2\varepsilon }}>1$ in the given range). Hence if $s$ is so large that $(1 - 2\varepsilon )\log s >x_0$, then \begin{align*} \log D & \le \pi ((1 - 2\varepsilon )\log s)\log ((1 - 2\varepsilon )\log s) \\ & \le \frac{{1 - \varepsilon }}{{1 - 2\varepsilon }}\frac{{(1 - 2\varepsilon )\log s}}{{\log ((1 - 2\varepsilon )\log s)}}\log ((1 - 2\varepsilon )\log s) \\ &= (1 - \varepsilon )\log s . \end{align*}