Let $\pi(x)$ denotes the number of primes less than or equal to $x$. Also suppose that for some fixed $N$ we have $\pi(x+y)\ge\pi(x)+\pi(y)$. The problem is,
Show that the equality in the above inequality cannot hold for all $x,y>3$ and $x+y\le N$ where $x$ and $y$ are both positive integers.
Suppose that there exists some $x_0$ and $y_0$ such that we have $$\pi(x_0)+\pi(y_0)=\pi(x_0+y_0)$$ Now if $x_0+y_0$ is composite then $\pi(x_0+y_0)=\pi(x_0+y_0-1)$. Now note that none of $x_0$ and $y_0$ can be prime because then we will have a contradiction. So the essence of all this is that if $x_0+y_0$ is composite then we also have a solution of the equation $$\pi(x+y-1)=\pi(x’)+\pi(y’)$$ such that $x’+y’=x+y-1$.
But I cannot solve the case when $x_0+y_0$ is a prime. Can anyone help me?
After going through the answer below, I think that I should elaborate once more the problem. What it says in disguise (so far I have understood) is the following,
Prove that if $p$ be an odd prime then there doesn't exist any solution to the equation, $$\pi(p-1)=\pi(x)+\pi(p-1-x)$$ for any integer $x$ where $p-x$ is a composite number.
Hope that this clarifies the problem more. If there is any problem (e.g., vagueness) still remaining, please let me know.