Lower bound for $\phi(n)$: Is $n/5 < \phi (n) < n$ for all $n > 1$?

Question

Is it true that :

$\frac {n}{5} < \phi (n) < n$ for all $n > 1$

where $\phi (n)$ is Euler's totient function .

Since $\phi(n)$ has maximum value when $n$ is a prime it follows that maximum value of $\phi(n)$ in term of $n$ is $n-1$ , therefore $\phi(n)< n$ for all $n$.

What is the best lower bound for $\phi(n)$?

Answer 1

The statement is false, with the first counterexample being 30030, for which $\phi(30030) = 5760 < \frac{30030}{5} = 6006$.

Also, if it's of any interest, all counterexamples below 10 million have at least 6 distinct prime divisors. For example, $30030 = 2 \times 3 \times 5 \times 7 \times 11 \times 13$.

Answer 2

Since $\phi(n)=n (1-\frac{1}{p_1}) \dots (1-\frac{1}{p_k})$ (where $n=p_1^{a_1} \dots p_k^{a_k}$), it is equivalent to asking if $(1-\frac{1}{p_1}) \dots (1-\frac{1}{p_k}) > \frac{1}{5}$, the minimal counterexample must be a product of first $k$ primes (because taking a smaller prime decreases the product); by direct calculation $2 \cdot 3 \cdot \dots \cdot 13 = 30030$ gives about 0.192. In general the product diverges to 0, which follows from this proof. So your inequality is not true even if you replace 5 with a larger, fixed number, and the smallest counterexample will be the product of $k$ first primes for some $k$.

Answer 3

As the other answers have pointed out, the answer is no, and this remains true if we replace $5$ with any larger integer. This raises the question, what is the optimal lower bound? The following theorem completely answers this:

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.

Remark: Note in particular that since $\log \log n\rightarrow \infty$ as $n$ grows large, we see that the result $\frac{n}{m}<\phi(n)$ cannot hold for any fixed integer $m$.

Proof: Consider $\mathcal{R}$, the set of all $n$ such that $m<n$ implies $\frac{\phi(n)}{n}<\frac {\phi(m)}{m}$. This set is then all of the "record breaking" $n$. If $n\in\mathcal{R}$ has $k$ prime factors, let $n^*$ be the product of the first $k$ prime factors. If $n\neq n^*$, then $n^*<n$ and $\frac{\phi(n^*)}{n^*}\leq \frac{\phi(n)}{n}$, which is impossible. Hence $\mathcal{R}$ consist entirely of $n$ of the form $n=\prod_{p\leq y}p$ for some $y$.

Now for $n\in \mathcal{R}$, we can choose $y$ so that $\log n=\sum_{p\leq y} \log p=\theta (y)$. Then using one of Mertens estimates we see that $$\frac{\phi(n)}{n}=\prod_{p\leq y}\left(1-\frac{1}{p}\right)=\frac{e^{-\gamma}}{\log y}+O\left(\frac{1}{(\log y)^2}\right).$$ Since $\log \log n =\log (\theta(y))=\log y+O(1)$ by Mertens estimates again, we have for $n\in\mathcal{R}$ $$\phi(n)=\frac{ne^{-\gamma}}{\log\log n}+O\left(\frac{1}{(\log \log n)^2}\right).$$ This along with the definition of $\mathcal{R}$ implies the desired result.

Edit: Added in the proof. This proof and statement of the result is Theorem 2.9 in Montgomery and Vaughn's multiplicative number theory.