[Math] Bounds of Euler’s totient function

Question

Conjecture :

Let $\phi(m)$ be Euler's totient function

$1 \leq \phi(m) \leq \lceil \frac{m-1}{2} \rceil ~~$ if $~~m~~$ is even

$\lceil \frac{m+1}{3} \rceil \leq\phi(m) \leq m-1 ~~$ if $~~m~~$ is odd

Proof :

Case when $ m$ is an even number :

Let us make substitution : $m=n-1$ , so:

$n=2^r \cdot q_1^{r_1} \cdot q_2^{r_2} \cdots q_k^{r_k}+1 \Rightarrow $

$\Rightarrow \phi(n-1)=\phi(2^r) \cdot \phi(q_1^{r_1})\cdots \phi(q_k^{r_k})=\phi(2^r) \cdot q_1^{r_1-1}(q_1-1)\cdots q_k^{r_k-1}(q_k-1)=$

$=\phi(2^r) \cdot q_1^{r_1} \frac{(q_1-1)}{q_1} \cdots q_k^{r_k} \frac{(q_k-1)}{q_k} < \phi(2^r) \cdot \frac{n-1}{2^r}=2^{r-1}\cdot \phi(2) \cdot \frac{n-1}{2^r}=\frac{n-1}{2}= \lceil \frac{n}{2}-1 \rceil \Rightarrow$

$\Rightarrow \phi(m) < \lceil \frac{m-1}{2} \rceil$

Question : How to prove the second part of the statement, (case when $m$ is an odd number) ?

Answer 1

Statement 2: The upper bound does hold by the multiplicative property of $\phi(n)$, however the lower bound is not true. In fact for any fixed integer $k$, $\frac{m}{k} \leq \phi(m)$ cannot hold for all $m$. A counter example is given by considering the integer $$M_x:=\prod_{p\leq x,\ p\neq 2} p.$$
As $x$ grows $\frac{\phi(M_x)}{M_x}$ becomes arbitrarily small.

The exact lower growth rate is given precisely by the following theorem:

Theorem: For all $n\geq 3$ we have $$\phi(n)\geq \frac{n}{e^{\gamma}\log \log n}+O\left(\frac{n}{(\log \log n)^2}\right),$$ where $\gamma$ is the Euler-Mascheroni Constant, and the above holds with equality infinitely often.

For more details, and a proof of the above, see this answer.