Prove the Inequality on $\pi$-function

Question

Prove that for each $y \geq 2$ , we have $\pi(x)+\pi(y)>\pi(x+y)$ for all sufficiently large $x$.

I tried searching in the Internet for quite a while. The best result that I have found is L. Panaitopol's result that if $a \in (0,1]$ and $x \geq y \geq ax$ , $x \geq e^{9a^{-2}}$ then $\pi(x)+\pi(y) > \pi(x+y)$. Here $\pi(x)$ is defined as the function which counts the number of primes not greater than $x$.

Is there any way to prove my question?

Recently I have been able to prove that,

For all $y \geq 6$ there exists a real number $M$ such that for all $x \geq M$ we will have $\pi(x)+\pi(y) > \pi(x+y)$.

Proof

For a proof of this inequality we use the inequality $\pi(ky)+\pi(y) > \pi((k+1)y)$. The objective is to find a lower bound for $k$ above which for all $k$ the inequality holds.

From the inequality $\dfrac{x}{\ln x-(1-\epsilon)}<\pi(x)<\dfrac{x}{\ln x-(1+\epsilon)}$ which holds for all $\epsilon>0$ and for all sufficiently large $x$ we get, $$k \left(\displaystyle \frac{\ln \left(1+\displaystyle\frac{1}{k}\right)-2\epsilon}{\ln ky-(1-\epsilon)} \right) \geq \left(\displaystyle\frac{2\epsilon-\ln (k+1)}{\ln y-(1-\epsilon)}\right)$$

Now we note that, $\displaystyle\frac{k}{\ln ky - (1-\epsilon)} \geq \displaystyle\frac{1}{\ln y - (1-\epsilon)}$ for all $y \geq 6$ and for all $k>1$ because the above inequality is implied by, $\left(\displaystyle\frac{y}{e}\right)^{k-1} \geq k$ which holds for all $k>1$ and for all $y \geq 2e$.

Hence we are left with proving that $\ln \left(1+\displaystyle\frac{1}{k}\right)-2\epsilon \geq 2\epsilon-\ln (k+1)$ which is equivalent to $(k+1)^2 \geq ke^{4\epsilon}$ and this indeed holds for all sufficiently large $k$.

Thus according to the previous argument the problem reduces to finding a suitable $\epsilon$ which may be said to be the $\sup$ of the set of all such $\epsilon$'s which satisfy the bounds.

But according to the answer given below, there should be some flaw in my approach. What am I missing?

Answer 1

Here is some related results according to R. Garunkštis' paper On some inequalities concerning $\pi(x)$. All other paper references are in R. Garunkštis' paper.

It is an open problem, called second Hardy–Littlewood conjecture $(1923)$, which states that $$\pi(x+y) \leq \pi(x) + \pi(y),$$ for all $x,y \geq 2$.

Schinzel and Sierpinski in $1958$ proved that $\pi(x+y) \leq \pi(x) + \pi(y)$ is true for all $2 \leq \min\{x,y\} \leq 146$. Gordon and Rodemich in $1998$ proved that $\pi(x+y) \leq \pi(x) + \pi(y)$ is true for all $2 \leq \min\{x,y\} \leq 1731$.

Dusart obtained in $1998$ that if $x \leq y \leq \frac 75 x \log x \log\log x$, then $\pi(x+y) \leq \pi(x) + \pi(y)$.

Answer 2

The inequality is surely wrong as written. Take $y = 2.9$, say, so that $\pi(y) = 1$. Then if $x$ is slightly less than a pair of twin primes, the inequality fails, and we expect there to be infinitely many twin primes. More generally, this inequality seems to fail for many values of $y$.

The question is asking whether for fixed $y \geq 2$, $\#\{p \leq y\} > \limsup_{x \to \infty} \#\{x < p \leq x + y\}$. One the one hand, this seems extremely likely, as primes are common among small numbers but rare among large numbers. On the other hand, likely is not the same as true; there could be occasional "clumps" of primes in a short interval.

To be more precise, write $p_n$ for the $n$-th prime. Then the question is asking whether for each positive integer $m$, the constant \[H_m := \liminf_{n \to \infty} p_{n + m} - p_n\] satisfies the inequality $H_m > p_{m + 1}$ (if this inequality were not true, so that there exists some subsequence $\{n_k\}$ for which $p_{n_k + m} - p_{n_k} < p_{m + 1}$ for all $k$, then take $x = p_{n_k}$ and $y = p_{m + 1} - 0.1$ to obtain the original question). Now $p_m \sim m \log m + m \log \log m - m$ by (inverting) the prime number theorem, whereas numerical evidence (see here) suggests that $H_m \approx m \log m + m$ (and indeed often seems to be less than this). To actually prove this is the case, however, would require the prime $k$-tuples conjecture, which is open.