A question on the inequality $\bigl(\pi(x+y)\bigr)^2<4\pi(x)\pi(y)$ From the answer of this post it seems highly probable that the following problem can be proved, > Show that for all sufficiently large $\min(x,y)$ we will have, $$\bigl(\pi(x+y)\bigr)^2<4\pi(x)\pi(y)$$Find the optimal lower bound of $\min(x,y)$ so that the above inequality holds. Here $\pi(x)$ denotes the number of primes less than or equal to $x$. However, the only thing that I have found till now is that when $x=y$, the inequality trivially holds. Note also that we have, $\bigl(\pi(x+y)\bigr)^2<4\bigl(\pi(x)\bigr)^2$ for all $x\ge 3$ and $x\ge y$. Can anyone suggest any method of showing it?

Is this not false? Assume contrariwise that there exists a lower bound $C$ such that the claim holds for all $x,y\ge C$. Let $y=C$. There are infinitely many prime gaps of length $C$, say the interval $[k!+2,k!+C+2]$ for any $k>C+2$. So if $x$ is the lower end of such an interval, we have $\pi(x+y)=\pi(x)$.

In that case your claim reads $$ \pi(x)^2<4\pi(x)\pi(C), $$ or, equivalently $$ \pi(x)<4\pi(C). $$ Because here $C$ is fixed, and $x$ can be made as large as we wish, this is absurd.