TITLE: Is the Euler phi function bounded below?
QUESTION [26 upvotes]: I am working on a question for my number theory class that asks:

Prove that for every integer $n \geq 1$, $\phi(n) \geq \frac{\sqrt{n}}{\sqrt{2}}$.

However, I was searching around Google, and on various websites I have found people explaining that the phi function has a defined upper bound, but no lower bound. Am I reading these sites incorrectly, or am I missing something in the problem itself?

REPLY [6 votes]: Firstly: the function $\phi(n)$ does not have the lower limit in the form of a straight line with a positive slope.
This follows from the formula 
$$\dfrac{\phi(n)}n = \prod\limits_{p\,|\, n}\dfrac{p-1}p,$$ formula, which illustrates non-increasing of $\dfrac{\phi(n)}n$.
Analysis of the formula leads to the following conclusions for :
$$\dfrac{\phi(np^k)}{np^k} = \begin{cases}\dfrac{\phi(n)}{n},\text{ if }p\,|\,n\\\dfrac{\phi(n)}{n}\dfrac{p-1}p,\text{ if }p\!\not|\:n\end{cases}$$
These findings can be illustrated by a simple table: 
So, not difficult to prove that 
$$\dfrac{\phi(n)}n \geq H(n),$$ where $H(n)$ is a step function
$$H(n) = \begin{cases} 
    \dfrac12,\text{ if } n \in [2,6)\\
    \dfrac13,\text{ if } n \in [6,30)\\
    \dots\\
    r_k,\text{ if } n \in [p_k\#, p_{k+1}\#)\\
    \dots,
\end{cases}$$
$\qquad p_k\# = \prod\limits_{i=1}^{k}p_i$ is primorial,
$$\left\{p_k\#\right\}=\{2,6,30,210,2310,30030,510510,96996900,223092870,6469693230,\dots\}$$
$$r_k = \dfrac1{p_k\#}\prod\limits_{i=1}^{k}(p_i-1),\dots,
\quad \{r_k\}=\left\{\dfrac12,\dfrac13,\dfrac4{15},\dfrac8{35},\dfrac{16}{77},\dfrac{192}{1001},\dfrac{3072}{17017},\dfrac{55396}{323323},\dfrac{110592}{676039},\dfrac{442368}{2800733},\dots\right\}$$
$$\{p_i\}=\{2,3,5,\dots\}.$$
This function is susceptible to various approximations.
As $p_i\geq i,$ then for not increasing consequence $R_k$ is valid ratio
$$r_k = \prod\limits_{i=1}^k\left(1-\dfrac1{p_i}\right)\geq \prod\limits_{i=1}^k\left(1-\dfrac1{i+1}\right) = \dfrac1{k+1}.$$
On the other hand, is not difficult to prove that $$p_k\#\geq k(k+1).$$
So for $(k-1)k < n \leq k(k+1):$
$$\sqrt{n}>k-1,$$
$$\dfrac{\phi(n)}n \geq\dfrac1{k(k+1)}\phi\left(k(k+1)\right) \geq \dfrac {\phi(p_k\#)}{p_k\#}=r_k \geq \dfrac1{k+1} > \dfrac1{\sqrt n +2}.$$
$$\boxed{\phi(n) \geq \dfrac n{\sqrt n+2}},$$