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 $0<a,b,c,d\leq 1$ and the constants $0<\alpha,\beta,\gamma\leq 1$ are very close to our upper limit $1$, for all real numbers $x<y$ with $L<x$ 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 $0<a,b,c,d\leq 1$ 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 $x<y$ for a suitable $L<x$? 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 $$\pi(x+y)-\pi(x)-\pi(y)\le$$ $$\operatorname{li}(x+y)- \operatorname{li}(x)- \operatorname{li}(y)+f(x)+f(y)+f(x+y)\le$$ $$ \frac 1{\log (x+y-1)}-2\operatorname{li}(1)+ f(x)+f(y)+f(x+y).$$