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Are there infinitely many solutions to the following inequalities (with $x\ne y$ and $x+y$ odd),

$$x+y>p_{\pi(x)}+p_{\pi(y)+1}\tag{1}$$and $$x+y>p_{\pi(x)+1}+p_{\pi(y)}\tag{2}$$where $\pi(x)$ denotes the number of primes less than or equal to $x$?

In one of my previous post it has been shown that the inequality, $$x+y<\dfrac{p_{\pi(x)}+p_{\pi(y)+1}+p_{\pi(x)+1}+p_{\pi(y)}}{2}$$fails infinitely often. So, one can expect that there are simultaneous solutions to $(1)$ and $(2)$ but I can't prove it. Can anyone help me?

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Yes. Let $p_k\ge p_{k-1}+4$ be any prime following a gap of at least 4 and let $x=p_k-1,y=p_k-2$. Then $$ x+y = 2p_k-3 \\ p_{\pi(x)}+p_{\pi(y)+1} = p_{\pi(x)+1}+p_{\pi(y)} = p_k+p_{k-1} \le 2p_k-4 $$

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