In terms of real parameter $a$ solve inequality:
$\log _{a} {x} + \mid a+\log _{a} {x} \mid \cdot \log _{\sqrt{x}} {a} \geq a\log _{x} {a}$
in set of real numbers
Hint:
Denote ${{\log }_{a}}x=t$, and we get: $\frac{{{t}^{2}}+2\left| t+a \right|-a}{t}\ge 0$ and just use that $\left| t+a \right|=\left\{ \begin{matrix}
t+a & t\ge -a \\
-t-a & t<a \\
\end{matrix} \right.$
Dear gobathegreat,
Sorry for spaming but you may want to create a contest section for the "Regional Olympiad - Federation of Bosnia and Herzegovina 2017" instead of posting too much threads. It'd be more convenience for other people
gobathegreat wrote:
In terms of real parameter $a$ solve inequality:
$\log _{a} {x} + \mid a+\log _{a} {x} \mid \cdot \log _{\sqrt{x}} {a} \geq a\log _{x} {a}$
in set of real numbers
https://artofproblemsolving.com/community/c6h1666857p10593427