UPDATE 2015.10.21: Thanks to Peter Mueller, I read from notes of G.J.O. Jameson at http://www.maths.lancs.ac.uk/~jameson/cyp.pdf on
cyclotomic polynomials of a sharper result, which indeed is simpler but also more challenging. I remove some of the challenge by interpreting some highlights here (hopefully without errors), but I recommend following the development of the notes as it proceeds in small but useful steps, with a certain degree of economy that takes ones breath away.
First, Jameson notes in 1.3 an inversion relation involving $\Phi_n(1/x)$
and real,nonzero $x$ that appears below.
Jameson also prepares in 1.12 to work with squarefree indices through
using $n_0=$rad$(n)$ and the identity $\Phi_n(x) = \Phi_{n_0}(x^{n/n_0})$. I modify and sketch a strict inequality (Lemma 1.19) which is used:
For $0 \lt x \lt 1, m, a,\ldots,b$ positive integers (so also $0 \lt x^{powers} \leq x$),
\begin{eqnarray*}
(1 -x^m)(1-x^{m+a})\ldots(1-x^{m+b}) & \geq &
(1 - x^m - x^{m+a} - \ldots - x^{m+b}) \\
& \gt & 1 - ( x^m + x^{m+1} + \ldots ) = 1 - x^m/(1-x) \\
\end{eqnarray*}
Then Jameson has 1.20, which I rewrite and restrict to squarefree integers $n$, as one actually gets better bounds/ranges for when $n$ is not squarefree.
1.20 (rewritten) Let $n>1$ be squarefree with $j=1$ if the number $k$ of distinct prime factors
of $n$ is an even number, and $j=-1$ if $k$ is odd. Let $0 \lt x \leq 1/2$. Then
$$1-x \lt \Phi_n(x)^j \lt 1.$$
Note when $n$ is 1, one has $\Phi_1(x)= x-1$ which is negative on the domain considered.
Using the inversion $\Phi_n(x)/x^{\phi(n)} = \Phi_n(1/x)$ when $n \gt 1$, this gives for $2 \leq x$
\begin{eqnarray*}
(x-1)/x && \lt \Phi_n(x)/x^{\phi(n)} \lt 1, && k=2m \\
1 && \lt \Phi_n(x)/x^{\phi(n)} \lt x/(x-1), && k=2m+1 . \\
\end{eqnarray*}
Using the relation for general non-squarefree indices, one can improve the $x$ in $1-x$ to $x$ to a fractional power, as well as extend the range a little. I am still working this part out. Even working out the statement using the inversion requires care. I think the results are both simple and challenging, and I am glad to share this on MathOverflow.
Jameson uses the tools carefully, working out the squarefree case in about half a page of elementary reasoning which I am still perusing. I am joyed. I'm also willing to buy Jameson two hot beverages. Peter Mueller can drop by
and ask me for a toasted bagel.
END UPDATE 2015.10.21
Gerhard "Wants To Stop Spinning Head" Paseman, 2015.10.19