A Problem from Gaokao 2016

Note that by changing the base of the logarithms, $a\log_bc - b\log_ac = \frac{\ln c}{\ln a \ln b} \left(a\ln a - b\ln b\right)$

Since $a,b > 1 > c > 0$ , we have $\ln c < 0 < \ln a, \ln b$ , so $\frac{\ln c}{\ln a \ln b} < 0.$

Also, since $1 < b < a$ we have $\left\{\begin{array}{c} 0 < b < a \\ 0 < \ln b < \ln a \end{array}\right\} \implies 0 < b \ln b < a \ln a \implies a\ln a - b\ln b > 0$

It follows that $a\log_bc - b\log_ac = \frac{\ln c}{\ln a \ln b} \left(a\ln a - b\ln b\right) < 0 \implies \boxed{a\log_bc < b\log_ac}$