Problem in deducing an identity related to Logarithm to the base e . I have a problem in deriving  this inequality about the  logarithm to base $e$ . 
Can someone please  give hints.

If $x>1$ then prove that $$\log x + \log\frac{x}2+ \log \frac {x} {2^2}  + \log \frac{x} {2^3}  +\dots < \log^2 (x) .$$

Can someone please tell how to prove this?
 A: Because
$$\begin{align}
   &\frac{\ln^2x}{\ln x+\ln\frac x2+\ln\frac x{2^2}+\cdots}\\
= &\frac{\ln^2x}{\ln\left(\displaystyle\prod_{i=0}^n\frac x{2^i}\right)}\\
= &\frac{\ln^2x}{\ln\left(\frac{x^n}{2^\frac{n(n-1)}2}\right)}\\
= &\frac{\ln^2x}{n\ln x-\frac{n^2-n}2\ln 2}\\
= &\frac{1}{\frac n{\ln x}-\frac{n^2-n}2\cdot\frac{\ln 2}{\ln^2x}}\\
= &\frac{1}{-\frac{\ln 2}{2\ln^2x}n^2+\left(\frac 1{\ln x}+\frac{\ln 2}{2\ln^2x}\right)n}
\end{align}$$
$\because\forall x>1,\ln x>0,\therefore-\frac{\ln 2}{2\ln^2x} <0,\therefore \lim\limits_{n\rightarrow\infty}-\frac{\ln 2}{2\ln^2x}n^2+\left(\frac 1{\ln x}+\frac{\ln 2}{2\ln^2x}\right)n =-\infty$.
$$\begin{align}
\therefore & \,\lim_{n\rightarrow\infty}\frac{1}{-\frac{\ln 2}{2\ln^2x}n^2+\left(\frac 1{\ln x}+\frac{\ln 2}{2\ln^2x}\right)n} =  \frac 1{-\infty}=0^-\\
\therefore & \,\frac{\ln^2x}{\ln x+\ln\frac x2+\ln\frac x{2^2}+\cdots}<0\\
\text{and}\,\because & \,\forall x>1,\ln^2x>0,\therefore \ln x+\ln\frac x2+\ln\frac x{2^2}+\cdots <0<\ln^2x
\end{align}$$
So $\ln x+\ln\frac x2+\ln\frac x{2^2}+\cdots<\ln^2x$
