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# Third CS's inequality

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.

1 year, 8 months ago

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- 1 year, 8 months ago

$${\phi(n+1)}^{p}-\phi(n^{p}) > n^{p-1}-{(n+1)}^{p-1}$$ 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.

- 1 year, 8 months ago

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

- 1 year, 8 months ago

then again subtracting $${(n+1)}^{p-1}$$ from both sides, we arrive at the required outcome.

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.

- 1 year, 8 months ago

Thanks for pointing that out ^^

- 1 year, 8 months ago

Just check the inequality , I have rephrased it. @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 ?

- 1 year, 8 months 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.

- 1 year, 8 months ago

Thanks ! I may ask for your help if I get stuck somewhere :D

- 1 year, 8 months ago

@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 ?

- 1 year, 8 months ago