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.

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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TopNewest@ZK LIn .do you think this stuff need more research or is it all nonsense ? I will be glad to here from you – Chinmay Sangawadekar · 1 year ago

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@Mehul Arora – Chinmay Sangawadekar · 1 year ago

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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. – Zk Lin · 1 year ago

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\(n>n-1\) , 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 – Chinmay Sangawadekar · 1 year ago

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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. – Zk Lin · 1 year ago

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– Chinmay Sangawadekar · 1 year ago

Thanks for pointing that out ^^Log in to reply

@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 ? – Chinmay Sangawadekar · 1 year ago

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

– Zk Lin · 1 year ago

This result seems pretty trivial to be published in a paper. The fact that \(\phi(p^{n})=p^{n}-p^{n-1}\) is well-known. If you like NT inequalities, this might serve as a good starting point for further explorations though. I am sure you will be able to discover something worth publishing if you keep up with your endeavor.Log in to reply

– Chinmay Sangawadekar · 1 year ago

Thanks ! I may ask for your help if I get stuck somewhere :DLog in to reply

– Chinmay Sangawadekar · 1 year ago

\(n\) is a prime and \(p\) is NLog in to reply

@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 ? – Chinmay Sangawadekar · 1 year ago

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