Second Hardy-Littlewood Conjecture theme

If Second Hardy-Littlewood Conjecture is true then we can claim that $\pi(x)-\pi(y)\leq \pi(x-y)$. Thus the conjecture gives an upper bound for the number of primes between $x$ and $y$. I have found some similar results that gives an upper bound for the number of primes in some of the papers (for example see this paper).

But my question is:

is there any conjecture, say of the form $\ \pi(x)-\pi(y)\geq a\pi(f(x,y)),\$ which holds for all sufficiently large $x$ and $y,$ and where $\ a\$ is a constant greater than $1$?

For a more clarification of the conjecture and consequences let us just observe that if $a=1$, $f(x,y)=\sqrt{x}-\sqrt{y}$ then we can have $\pi(x)-\pi(y)\geq \pi \left(\sqrt{x}-\sqrt{y}\right)$. Now putting $x=(n+1)^2$ and $y=n^2$ we see that a consequence of the inequality is Legendre Conjecture.

Basically, most of the conjectures regarding the distribution of primes between two numbers asks us to prove that there is at least one prime between the two said numbers. In this aspect I think that this type of inequality may be particularly helpful.

In a similar manner some other conjectures regarding the distribution of prime numbers between may be tackled very effectively if a good expression for $f$ be found out.

• Sure. Bertrand's postulate is the case $f(x,y) = 2$ if $x > 2y$ and zero otherwise. – S. Carnahan Oct 23 '14 at 8:17
• Strange question. Take $a=10^{12}$ and $f(x,y)=x^{\frac 13}-y^{\frac 13}$. Huxley's version of the prime number theorem in short intervals, gives the inequality for all large $x$ and $y$. Not clear to me why this has any more interest than Huxley's theorem! – Lucia Oct 26 '14 at 14:17