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New types of inequalities.

Prove/disprove the following inequalities.

If \(M\) and \(k\) are natural numbers, then

\[1) \sum_{n=1}^{M} \phi(n) \geq \frac{1}{\sum_{n=1}^{M} \phi (n)}\]

\[2)\sum_{n=1}^{M} \phi(n) \geq \sum_{n=1}^{M} \phi(\phi(n))\]

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

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Note by Chinmay Sangawadekar
1 year, 8 months ago

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  1. since the sum on the LHS is ≥1, it is ≥ its reciprocal.

  2. Lemma:\(\phi(n^k)≥\phi(n)^k\)

proof:\(\phi(n^k)=n^{k-1}\phi(n)≥\phi^k(n)\to n^{k-1}≥\phi^{k-1}(n)\to n≥\phi(n)\) which we know is true. the result follows

3.\(n≥\phi(n)\) let \(n=\phi(x)\) the \(\phi(x)≥\phi(\phi(x))\) and the result follows.

Aareyan Manzoor - 1 year, 8 months ago

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Darn I worked so hard to type the whole solution.

Thank god I started from the bottom, otherwise I would've worked harder, only to receive nothing :P

Mehul Arora - 1 year, 8 months ago

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right, exactly. The only non-trivial one is \(\phi(n^k)=n^{k-1}\phi(n)\)

Otto Bretscher - 1 year, 8 months ago

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Q3. We will prove that \(\phi (n) > \phi (\phi (n))\)

\(\phi (n) = n \times (1- \dfrac {1}{a_1}) \times (1- \dfrac {1}{a_2}) \times \cdots \times \dfrac {1}{a_k}\) where \(n = a_1 ^{p_1} a_2 ^{p_2}\cdots a_k ^{p_k}\) where \(a_1,a_2...., a_k\) are prime numbers, and \(p_1,p_2..., p_k\) are positive integers.

Now, \(\phi (\phi (n)) = \phi (n \times (1- \dfrac {1}{a_1}) \times (1- \dfrac {1}{a_2}) \times \cdots \times \dfrac {1}{a_k}))\)

We observe that \(\phi (n) < n\) , because it is multiplied by fractions less than 1.

Thus , we show that \(\phi (\phi (n)) \leq \phi (n)\) , thus the inequality holds true.

Mehul Arora - 1 year, 8 months ago

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Nicely done Mehul !

Chinmay Sangawadekar - 1 year, 8 months ago

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Thanks Chinmay! :D

I was about to write the other proofs as well, but Aareyan beat me ;)

Mehul Arora - 1 year, 8 months ago

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@Mehul Arora oh , want some more ? actually I didn't know that 2nd one was a lemma , ....

Chinmay Sangawadekar - 1 year, 8 months ago

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Small correction : \(\phi(n)\leq n\) since you have equality for \(n=1\)

Otto Bretscher - 1 year, 8 months ago

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Oops yeah, thanks :D

Mehul Arora - 1 year, 8 months ago

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I thought about them during my morning shower, and they all appear to be true. Want any proofs? They are all one liners...

Otto Bretscher - 1 year, 8 months ago

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Morning shower is the best place to think abt maths , ;) , Actually I thought abt them and their proofs , in my morning shower

Chinmay Sangawadekar - 1 year, 8 months ago

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Comment deleted Mar 02, 2016

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Post them*

I have* :P

Mehul Arora - 1 year, 8 months ago

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