Riemann $\zeta$ and Chebyshev's estimates

How about this for a start? It shows $\limsup_{x \to \infty} \frac{\pi(x)}{x/\log x} \ge 1$.

For $\sigma > 1$, $$\log \zeta(s) = \log \prod_p (1-p^{-s})^{-1} = \sum_p \sum_{m \ge 1} \frac{1}{mp^{ms}} = \sum_p \frac{1}{p^s}+O_{s \to 1}(1).$$ Now let $s=1+\epsilon$ for small $\epsilon > 0$. Then $$\log \frac{1}{\epsilon} \sim \log \zeta(s) = \sum_p \frac{1}{p^s} = \lim_{x \to \infty} \sum_{p \le x} \frac{1}{p^s} = \lim_{x \to \infty} \frac{\pi(x)}{x^s}+s\int_1^x \frac{\pi(t)}{t^{s+1}}dt = (1+\epsilon)\int_1^\infty \frac{\pi(t)}{t^{2+\epsilon}}dt.$$

Now if there were some $\delta > 0$ so that $\pi(t) \le (1-\delta)\frac{t}{\log t}$ for all large $t$, then $$1 = \lim_{\epsilon \downarrow 0} \frac{\log(1/\epsilon)}{\log(1/\epsilon)} = \lim_{\epsilon \downarrow 0} \frac{1}{\log(1/\epsilon)}\int_1^\infty \frac{\pi(t)}{t^{2+\epsilon}}dt \le \lim_{\epsilon \downarrow 0} \frac{1}{\log(1/\epsilon)}(1-\delta)\int_2^\infty \frac{1}{t^{1+\epsilon}\log(t)}dt = 1-\delta,$$ a contradiction.