I was thinking about the Second Hardy-Littlewood conjecture for quite sometime (some of my posts are related to this). In one of my [earlier post](https://math.stackexchange.com/questions/959502/partial-proof-of-second-hardy-littlewood-conjecture-modified) I conjectured that the inequality ($\pi(x)$ denotes the prime counting function), $$\pi(x)+\pi(y)>2\pi\left(\dfrac{x+y}{2}\right)$$ is true. However, R. Israel showed that this inequality falls infinitely often. Basing upon the counter example that he gave I put forward the following two conjectures,



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> **Conjecture 1** $\pi(x)+\pi(y)\le2\pi\left(\dfrac{x+y}{2}\right)$ for all sufficiently large $x$ and $y$.
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It seems that the former conjecture is true if the quantity $\dfrac{x}{y}$ be sufficiently large (assume $x\ge y$). 


So my question is that,



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> Is there any proof of **Conjecture 1**?
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Assuming there is no such proof at present can anyone suggest me some relevant literature which shed some light on proving this conjecture?



A conjecture on the prime counting function
Look for a large gap in the distribution of primes. For this conjecture, the gap between $n!+2$ and $n!+n$ will suffice. Set $y = n!+2$ (which is composite) and set $m$ (which will be $\frac{x+y}{2}$, so $x$ will be $2m-y$ eventually) to be the largest composite so that there are no primes between $m$ and $y$. Then there will be primes between $x$ and $m$ and none between $m$ and $y$. 


So $\pi(x) + \pi(y) = \pi(x) + \pi\left(\frac{x+y}{2}\right) > 2 \pi\left(\frac{x+y}{2}\right)$. So your conjecture will fail for infinitely many pairs $x$ and $y$. 


Gerhard "This Belongs On Another Forum" Paseman, 2015.06.30


