How do we show that For $x,y \ge 0$ real numbers, there exists a constant C suchthat: $\pi(x+y)-\pi(y) \le \frac{1}{3}x+C$ Where $\pi(.)$ denotes thes prime counting function, is true?
the hint is to sieve n with $y< n \le x+y $:
$\pi (x+y) \le 1+ \sum _{n \le x+y} 1+1-1 - \sum_{2|n}1 - \sum_{3|n}1 + \sum_{6|n}1 + \sum_{n\le x+y} 1 = $ $1+ \sum _{n \le x+y} 1+1-1 - \sum_{2|n}1 - \sum_{3|n}1 + \sum_{6|n}1 + [x+y]$
because: $1\le n = dm \le x+y \Leftrightarrow \frac{1}{d}\le m \le \frac{x+y}{y} $
so: $\sum_{n\le x+y , d|n}1 = [\frac{x+y}{d}]$
then that gives: $\pi (x+y) < 1+ [x+y] - [\frac{x+y}{2}] - [\frac{x+y}{3}] + [\frac{x+y}{6}]$
so that will give: $\pi (x+y) < \frac{x+y}{3} + 3$ but also we get : $\pi(y) < \frac{y}{3} + 3$ so for any constant $C\ge 0$ it will surely hold that:
$\pi(x+y) - \pi(y) < \frac{x}{3} \le \frac{x}{3} + C$
Is this correct?