Prime counting inequality: [closed]
Assuming that:
$$\pi(x) \sim \frac{x}{log (x)}; \pi(y) \sim \frac{y}{log (y)}$$
Prove that:
$$\pi(x) - \pi(y) \le \frac{ψ(x)}{log(y)}$$
Where
$$ψ(x) \le \pi(x) *ln(x)$$ I don't know where to start...
Solution 1:
$$ \pi(x)-\pi(y)=\sum_{y<p\leqslant x}1\leqslant\sum_{y<p\leqslant x}\frac{\log p}{\log y} \leqslant\frac{\vartheta(x)}{\log y}$$ Now, using $\displaystyle \psi(x)=\sum_{n=1}^{+\infty}\vartheta(x^{1/n})\geqslant \vartheta(x)$, we have $\displaystyle\pi(x)-\pi(y)\leqslant\frac{\psi(x)}{\log y}$.