Prove/Disprove the following inequality.

\[ \sum_{cyc}\sum_{k=1}^{M}\phi(a_1^k)\geq\sum_{cyc}\sum_{k=1}^{M} \phi (a_1)^k\]

For all \(0\leq a_1 , a_2 , a_3 , ....., a_n \leq M\) and \(M\) is any natural number.

I have made the inequality more generalized.

**Notation**: \(\phi(\cdot) \) denotes the Euler's totient function.

You may also see Second CS's inequality

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## Comments

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TopNewestNot quite sure about this, correct me if I make a mistake somewhere.

Let \(p\) denotes an arbitrary prime number and let \(k,b \geq 1 \in \mathbb{N}\)

\(p^{k-1} \geq (p-1)^{k-1}\)

\(p^{-1} \geq p^{-k}(p-1)^{k-1}\)

\(p^{kb-1}(p-1) \geq p^{kb-k}(p-1)^{k}\)

\(p^{kb}-p^{kb-1} \geq {(p^{b}-p^{b-1})}^{k}\)

\(\phi(p^{kb}) \geq {\phi(p^{b})}^{k}\), noting that \(\phi(p^{n})=p^{n}-p^{n-1}\)

Let \(a={p_{1}}^{k_{1}}{p_{2}}^{k_{2}}\cdots{p_{n}}^{k_{n}}\).

From the result above, we have

\(\phi({p_{1}}^{kb_{1}})\phi({p_{2}}^{kb_{2}})\cdots\phi({p_{n}}^{kb_{n}}) \geq \phi{({p_{1}}^{b_{1}})}^{k}\phi{({p_{2}}^{b_{2}})}^{k}\cdots\phi{({p_{n}}^{b_{n}})}^{k}\)

\(\phi(a^{k}) \geq {\phi(a)}^{k}\)

Therefore, the inequality holds true, with equality holds only when \(a,b,c \in \{0,1\}\) or \(M=1\).

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Comment deleted Mar 04, 2016

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Done.

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Nicely done ! I don't think that there will be any mistake...@ZK LIn

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Thanks! Glad to know that.

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@Otto Bretscher What is your opinion ?

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first batch already, question 2?

Didn't we do this inequality in theIt is good to know the equation \(\phi(n^k)=n^{k-1}\phi(n)\); this inequality is an immediate consequence.

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Comment deleted Mar 04, 2016

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@Otto Bretscher

Do you think that such inequalities will be useful for ay theorem proofs or problems ?Log in to reply

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For \(k>1\),

\[\phi(a^k)=a^{k-1}\phi(a)\]

Since

\[\phi(a)<a\]

\[\phi(a)^k<a^{k-1} \phi(a)=\phi(a^k)\]

For k=1, the result \(\phi(a^k)=\phi(a)\) is trivial.

The result above follows.

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Exactly ! :)

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Comment deleted Mar 02, 2016

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Bro, I see that you have studied all these awesome stuffs!

Can you please suggest me resources for studying them?Thanks!

(I mean modular arithmetic)

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You can see that the inequality is proved , I had proved this only for arbitary peimes , and I am searching a more general proof , as you xan see the generalized inequality ... I will be glad to hear your opinions related to this new NT inequality.. @Harsh Shrivastava

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Comment deleted Apr 30, 2017

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People like @ZK LIn @MS HT @gurido cuong are real inequality gods.

Please suggest us how to become good at inequalities ?

Thnx!

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However, since you ask for advice, I will do my best to provide them.The "advice" below comes from my personal experience and might not work for everyone, so take it with a pinch of salt.

I initially start studying inequalities seriously from the excellent note Mildorf "Olympiad Inequalities", which I believe is available somewhere on the web (AoPS maybe?). You might find that helpful.

A more elementary version of this is Siamin Riasat "Basics in Olympiad Inequalities". When I am a complete newbie, I rely on it heavily and tried (unsuccessfully) to solve all the problems. Easy as the questions are, they provide a good chance for you to apply the inequalities introduced.

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Here

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BTW your joke was awesome.I am rolling over floor laughing.I appreciate efforts that you put in making this joke.You must consider a career as a stand-up comedian.Hats off to you!

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