# Inequalities $\pi(x^a+y^b)^\alpha\leq \pi(x^c)^\beta+\pi(y^d)^\gamma$ involving the prime-counting function, where the constants are very close to $1$

Let $$\pi(x)$$ be the prime-counting function, I'm curious about if a suitable variant of the second Hardy–Littlewood conjecture (this corresponding Wikipedia)

$$\pi(x^a+y^b)^{\alpha}\leq\pi(x^c)^{\beta}+\pi(y^d)^{\gamma}\tag{1}$$ can be proved, where the constants $$0 and the constants $$0<\alpha,\beta,\gamma\leq 1$$ are very close to our upper limit $$1$$, for all real numbers $$x with $$L for a (yours) suitable choice of a constant $$L$$.

Question. Is it possible to prove any statement of the type $$(1)$$ under the cited requirements, for constants $$0 and constants $$0<\alpha,\beta,\gamma\leq 1$$ all these (all together/ simultaneously) very close to $$1$$, for all real numbers $$x$$ and $$y$$ such that $$x for a suitable $$L? Many thanks.

I don't know if this type of proposals $$(1)$$ are in the literature, or are essentially the same than the original second Hardy–Littlewood conjecture, when we require that those constants are very close to $$1$$.

If there is relevant literature answer my question as a reference request and I try to search and read those statements from the literature.

## References:

[1] G. H. Hardy and J. E. Littlewood, Some problems of ‘Partitio numerorum’ III: On the expression of a number as a sum of primes, Acta Math. (44): 1–70 (1923).

I guess these constants lead to a wrong way to the second Hardy–Littlewood conjecture. Indeed, according to Wikipedia, for $$x\ge L_0=355991$$ we have $$\frac x{\log x} \left( 1+\frac 1{\log x}\right)<\pi(x)< \frac x{\log x} \left( 1+\frac 1{\log x}+\frac {2.51}{(\log x)^2}\right).$$ Then for $$x,y\ge L_0$$ the difference between the left-hand side and the right-hand side of the conjecture experession is not so big. Namely, $$\pi(x+y)-\pi(x)-\pi(y)\le$$ $$\frac {x+y}{\log (x+y)} \left( 1+\frac 1{\log (x+y)}+\frac {2.51}{(\log (x+y))^2}\right)- \frac x{\log x} \left( 1+\frac 1{\log x}\right)-\frac y{\log y} \left( 1+\frac 1{\log y}\right)\le$$ $$\frac {x+y}{\log (x+y)} \left( 1+\frac 1{\log (x+y)}+\frac {2.51}{(\log (x+y))^2}\right)- \frac x{\log (x+y)} \left( 1+\frac 1{\log (x+y)}\right)-\frac y{\log (x+y)} \left( 1+\frac 1{\log (x+y)}\right)=$$ $$\frac {2.51(x+y)}{(\log (x+y))^3}.$$

Thus I guess that if $$\max\{a,b\}\alpha<\min\{c\beta, d\gamma\}$$ then the right-hand side of the expression from your question grows asymptotically faster than the left-hand side and so the inequality holds for sufficiently big $$L$$.

So a way to the conjecture is to obtain better upper bounds for $$\pi(x+y)-\pi(x)-\pi(y)$$ than $$\frac {2.51(x+y)}{(\log (x+y))^3}$$ and I guess advances in this direction can be found in papers related to the conjecture.

Remark that Trudgian’s upper bound for the difference between $$\pi(x)$$ and $$\operatorname{li}(x)=\int_0^x \frac {dt}{\ln t}$$, $$|\pi(x)- \operatorname{li}(x)|\le f(x)= 0.2795\frac {x}{(\log x)^{3/4}}\exp\left(-\sqrt{\frac{\log x}{6.455}} \right)$$ for $$x\ge 229$$ implies that for $$x,y\ge 229$$ we have $$\pi(x+y)-\pi(x)-\pi(y)\le$$ $$\operatorname{li}(x+y)- \operatorname{li}(x)- \operatorname{li}(y)+f(x)+f(y)+f(x+y)\le$$ $$\frac 1{\log (x+y-1)}-2\operatorname{li}(1)+ f(x)+f(y)+f(x+y).$$

• Many thanks for your excellent answer, when I can it I'm going to study it. Commented Nov 8, 2020 at 16:33