<s>[INST] How can we show that $\pi (x+y) - \pi(y) \le \frac{1}{3} x + C$ using the sieve of eratosthenes? 
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? 
 [/INST] You are asking us to show that your inequality holds for any $x, y \ge 0$ and for any constant $C$, which is obviously false. Perhaps this is what you mean:  
Show that there exists a constant $C$ such that for any real numbers $x, y \ge 0$, $\pi(x+y)-\pi(y) \le \frac{1}{3}x+C$.
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