<s>[INST] 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
 [/INST] 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.
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