Q:

A proof of the approximate expression about totient summatory function

I'm a high school student in Korea.
I am preparing for a presentation. so I prove an approximate expression about totient summatory function , but I'm not sure that the proof is correct.
If the proof is incorrect , please tell me what is incorrect. 
The following is the proof :
(Prove  $ \Phi \left ( x \right ) \sim  \frac{x^{2}}{2\zeta \left ( 2 \right )}$)
$\Phi \left ( x \right )= \sum_{k<x}\varphi \left ( k \right ) = \sum_{dm<x}m\mu (d)$         
$=\sum_{d<x}\mu\left ( d \right )\sum_{m<x/d}m$
$=\sum_{d<x}\mu\left ( d \right ) \cdot \frac{1}{2} \cdot (\left [ \frac{x}{d} \right ]^{2} + \left [ \frac{x}{d} \right ] )$
$\sum_{d<x}\mu\left ( d \right )\left [\frac{x}{d}  \right ] = 1 $  so  $\Phi \left ( x \right ) - \frac{1}{2} = \sum_{d<x}\mu\left ( d \right ) \cdot \frac{1}{2} \cdot \left [ \frac{x}{d} \right ]^{2}  $
$\mu\left ( d \right )\frac{1}{2}(\frac{x^{2} - 2xd + d^{2} }{d^{2}}) < \mu\left ( d \right )\frac{1}{2}\left [ \frac{x}{d} \right ]^{2}  < \mu\left ( d \right )\frac{1}{2}(\frac{x^{2} }{d^{2}})
$
(because $\frac{x}{d} - 1 <  \left [ \frac{x}{d} \right ] \leq \frac{x}{d}$)
so
$\sum_{d<x}\mu\left ( d \right )\cdot\frac{1}{2}\cdot(\frac{x^{2} - 2xd + d^{2} }{d^{2}}) <\Phi \left ( x \right ) - \frac{1}{2} < \sum_{d<x}\mu\left ( d \right )\cdot\frac{1}{2}\cdot\frac{x^{2}  }{d^{2}}   $
$\Rightarrow \frac{x^{2}}{2}\sum_{d<x}\frac{\mu\left ( d \right )}{d^{2}} - x \sum_{d<x}\frac{\mu\left ( d \right ) }{d} + \frac{1}{2} \sum_{d<x} \mu \left ( d \right  )
 <\Phi \left ( x \right ) <\frac{x^{2}}{2}\sum_{d<x}\frac{\mu\left ( d \right )}{d^{2}} $
we know $ \left | \sum_{d<x}\frac{\mu\left ( d \right ) }{d}  \right |\leq 1$ and $\sum_{k=1}^{\infty}\frac{\mu \left ( s \right )}{n^{-s}} = \frac{1}{\zeta \left ( s \right )}$
Finally,
$\lim_{x->\infty }\frac{\Phi \left ( x \right )}{x^{2}} = \frac{1}{2\zeta \left ( 2 \right )} $
$\Rightarrow \Phi \left ( x \right ) \sim  \frac{x^{2}}{2\zeta \left ( 2 \right )}$

A:

Since $\varphi(n) = \sum_{d | n} \mu(d) \frac{n}{d}$ and $\sum_{d < x} \mu(d) \lfloor x/d \rfloor = 1$ $$\sum_{n < x} \varphi(n) = \sum_{n < x}\sum_{d | n} \mu(d) \frac{n}{d}=\sum_{md < x} \mu(d) m$$
$$=\frac{1}{2} \sum_{d<x}\mu(d)  (\left\lfloor \frac{x}{d} \right\rfloor^2 + \left\lfloor \frac{x}{d} \right\rfloor )=\frac{1}{2} +\frac{1}{2} \sum_{d<x}\mu(d)  \left\lfloor \frac{x}{d} \right\rfloor^2$$
$\frac12\left|\left\lfloor \frac{x}{d}  \right\rfloor^{2}-\frac{x^2}{d^2}\right| < 1+\frac{x}{d}$ so that $$\frac12\left|\sum_{d<x}\mu( d)   \left(\left\lfloor \frac{x}{d}  \right\rfloor^{2}-\frac{x^2}{d^2}\right)\right| <\sum_{d<x} (1+\frac{x}{d})< x+x \ln (x+1)$$
Also
$$\left|\sum_{d < x} \frac{\mu(d)}{d^2} -\frac{1}{\zeta(2)}\right| < \sum_{d \ge x} \frac{1}{d^2} < \frac{1}{x-1}$$
Therefore
$$\sum_{n < x} \varphi(n) \sim \frac12\sum_{d<x}\mu( d)  \left\lfloor \frac{x}{d}  \right\rfloor^{2} \sim \frac12\sum_{d<x}\mu( d) \frac{x^2}{d^2} \sim \frac{x^2}{2\zeta(2)}$$

