Let $02$ 
Let $0<a<b<1$ and $E=\log_{a}{\frac{2ab}{a+b}} + \log_{b}{\frac{2ab}{a+b}}$


Prove that $E>2$

I tried to change both log bases to $\frac{2ab}{a+b}$ and I got $E=\frac{1}{\log_{\frac{2ab}{a+b}}{a}} + \frac{1}{\log_{\frac{2ab}{a+b}}{b}}$.
Also I noticed that $\frac{2ab}{a+b} < 1$ because $\frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b}$ which is the harmonic mean of $a$ and $b$ and since harmonic mean is always smaller than $max(a,b)$, $\frac{2ab}{a+b}$ will be smaller than 1.
Since both the base and the argument of  $\log_{\frac{2ab}{a+b}}{a}$ and $\log_{\frac{2ab}{a+b}}{b}$
are smaller than 1, the logarithms will be positive, but I do not know if they will be greater than 1 or smaller than 1.
Any tips ? Thanks in  advance !
 A: Bringing to $\ln$, we have
$$\begin{align}
E
&=\log_{a}{\frac{2ab}{a+b}} + \log_{b}{\frac{2ab}{a+b}} \\
&= \frac{\ln\frac{2ab}{a+b}}{\ln a} + \frac{\ln\frac{2ab}{a+b}}{\ln b} && \left(1\right) \\
&= \ln\frac{2ab}{a+b}\left(\frac{1}{\ln a} + \frac{1}{\ln b}\right) \\
&> \ln\sqrt{ab}\left(\frac{1}{\ln a} + \frac{1}{\ln b}\right) && \left(2\right)\\
&= \frac{1}{2}(\ln a + \ln b)\left(\frac{1}{\ln a} + \frac{1}{\ln b}\right) \\
&= \frac12 \left(2 + \frac{\ln a}{\ln b} + \frac{\ln b}{\ln a}\right)\\
&> \frac 12\left(2 + 2\right) && (3)\\
&> 2
\end{align}$$
Explanation:
$(1)$ $\displaystyle\because\log_cd = \frac{\ln d}{\ln c}$
$(2)$ $\because\displaystyle \frac{2ab}{a+b} < \frac{2ab}{2\sqrt{ab}} = \sqrt{ab}$ (Thanks to @Angelo) and $ c_1d > c_2d$ for $c_1 < c_2 < 0,~ d <0$
$(3)$ $\because \displaystyle\frac{\ln a}{\ln b} + \frac{\ln b}{\ln a} \gt 2\sqrt{\frac{\ln a}{\ln b} \cdot \frac{\ln b}{\ln a}} = 2$
A: We have $$E=\frac{2\log H(a,b)}{H(\log a,\log b)}$$ where $H$ denotes the harmonic mean. Showing $E>2$ is equivalent to showing $$\log H(a,b)>H(\log a,\log b)$$ for $a<b$ which is true by Jensen's inequality.
