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

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 $\phi(n)$?

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 $k$.

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.

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, $30030 = 2 \times 3 \times 5 \times 7 \times 11 \times 13$.