Prove/Disprove the following inequality:

\[\prod_{n \text{ prime}} (n+1)^p > \prod_{n \text{ prime}} \phi(n^p) + n^{p-1}\]

See Part 1 and Part 2 on my Brilliant profile.

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## Comments

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TopNewest@Mehul Arora

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\({\phi(n+1)}^{p}-\phi(n^{p}) > n^{p-1}-{(n+1)}^{p-1}\) doesn't hold for all prime \(n\) and \(p \in \mathbb N\).

Suppose \(p=1, n=7\), then R.H.S.\(=0\), while L.H.S. \(=\phi(8)-\phi(7)=-2\), an instance where the inequality fails to hold.

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I just tried to do like this ,

\(n>n-1\) , so \((n+1)^p >n^p\) then subtracting \(n^{p-1}\) from both sides we,get , \((n+1)^p > \phi(n^p) + n^{p-1}\) , then again subtracting \((n+1)^{p-1}\) from both sides , we arrive at required outcome .pls do check and point out if there is any mistake. @ZK LIn

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This is wrong since \(n+1\) is not a prime. \(\phi(p^{n})=p^{n}-p^{n-1}\) is true only when \(p\) is a prime number.

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@ZK LIn , do you think that this concept of NT ineq. Is new ? can I search more of such and classify them , and present a paper ?

Just check the inequality , I have rephrased it.Log in to reply

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@Otto Bretscher can we classify such number therotic inequalities into a new group ? Though they do not have practical use but still just to classify ?

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