How prove this $\ln{(x+\sqrt{x^2+1})}<\frac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$ let $0<a<1,x<0$,show that
$$\ln{(x+\sqrt{x^2+1})}<\dfrac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$$
My idea:
$$\Longleftrightarrow \ln{(\sqrt{x^2+1}+x)}<\dfrac{x(a^x-1)\ln{a}}{(a^x+1)\ln{(\sqrt{x^2+1}-x)}}$$
Then following I fell very ugly.Thank you
 A: If $x<0$, then $x+\sqrt{x^2+1}<x+\sqrt{x^2-2x+1}=x+|x-1|=|x-1|-|x|\leq1$, so
$$\ln(x+\sqrt{x^2+1})<0$$
and $\sqrt{x^2+1}-x>1+0=1$, so
$$\ln(\sqrt{x^2+1}-x)>0$$
Additionally if $0<a<1$, then $a^x-1>0$, $a^x+1>0$ and $\ln a<0$, therefore
$$\frac{x(a^x-1)\ln a}{(a^x+1)\ln(\sqrt{x^2+1}-x)}=\frac{-+-}{++}>0$$
So the inequality holds quite trivially.
A: $$\ln{(x+\sqrt{x^2+1})}<\dfrac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$$
Rewrite the inequality in terms of hyperbolic functions:
1.) $\ln{(x+\sqrt{x^2+1})}=\sinh^{-1}x$
2.) $\frac{1}{\log_{a}{(\sqrt{x^2+1}-x)}}=-\frac{\log{a}}{\sinh^{-1}x}$
3.) $\frac{a^x-1}{a^x+1}=\tanh{\left(\frac{x\log{a}}{2}\right)}$
Using the three identities above, the inequality becomes:
$$\sinh^{-1}x<-\frac{x\log{a}}{\sinh^{-1}x}\tanh{\left(\frac{x\log{a}}{2}\right)}$$
For $x<0$ and $0<a<1$, 
$$\sinh^{-1}x < 0,\\
0<-x,\\
0<\frac{\log a}{\sinh^{-1}x},\\
0<\tanh{\left(\frac{x\log{a}}{2}\right)}.$$
Hence, 
$$\sinh^{-1}x<0<-\frac{x\log{a}}{\sinh^{-1}x}\tanh{\left(\frac{x\log{a}}{2}\right)}$$
