15.12. For which real numbers $x$ does the inequality
$$
2 \log _{x}\left(\frac{a+b}{2}\right) \leq \log _{x} a+\log _{x} b
$$
hold for all positive numbers $a$ and $b$ ?
15.12. $0<x<1$
The domain of the variable $x$ in the inequality is $(0,1) \cup(1,+\infty)$. Use properties of logarithms, the inequality can be rewritten into
$$
\log _{x}\left(\frac{a+b}{2}\right) \leq \log _{x} \sqrt{a b} .
$$

By the AM-GM Inequality, we know that
$$
\frac{a+b}{2} \geq \sqrt{a b}
$$

It follows that the function $\log _{x}$ is decreasing, which only happens when $0<x<1$.