How can I show that
$\displaystyle
\phi(n) \geq \sqrt{n/2}.
$
Write
$\displaystyle
n= {p_1}^n_1...{p_r}^n_r.
$
Then
$\displaystyle
\frac {\phi({p_1}^n_1)}{\sqrt {n}} = \frac{(p_1-1)p^{n_1-1}}{p_1^{n_1/2}}
$
then what?
How can I show that
$\displaystyle
\phi(n) \geq \sqrt{n/2}.
$
Write
$\displaystyle
n= {p_1}^n_1...{p_r}^n_r.
$
Then
$\displaystyle
\frac {\phi({p_1}^n_1)}{\sqrt {n}} = \frac{(p_1-1)p^{n_1-1}}{p_1^{n_1/2}}
$
then what?
First we must have $\displaystyle n\geq{2}$ for the ineuqality to hold
You'll need these 2 inequalities (by $\displaystyle p$ I mean prime):
$\displaystyle
\left\{ \begin{gathered}
\phi \left( {2^k } \right) = 2^{k - 1} \geqslant \sqrt {\tfrac{{2^k }}
{2}} = 2^{\tfrac{{k - 1}}
{2}} \hfill \\
\phi \left( {p^k } \right) = p^{k - 1} \cdot \left( {p - 1} \right) \geqslant \sqrt {p^k }{\text{ if }}p > 2 \hfill \\
\end{gathered} \right.
$ for $\displaystyle k\geq{1}$
For the second:
$\displaystyle
p - \sqrt p + \tfrac{1}
{4} = \left( {\sqrt{p} - \tfrac{1}
{2}} \right)^2 \geqslant \left( {\sqrt{3} - \tfrac{1}
{2}} \right)^2 \geqslant \tfrac{5}
{4}\therefore p - 1 \geqslant \sqrt p \geqslant p^{1 - \tfrac{k}
{2}}
$ $\displaystyle
\Rightarrow p - 1 \geqslant p^{1 - \tfrac{k}
{2}} $
Now your inequality follows easily from the fact that $\displaystyle \phi$ is multiplicative.
In fact, you can see from the inequalities above -check the equality conditions- that your inequality is strict for all $\displaystyle n>2$