9. It is given that $x$ and $y$ are two real numbers such that $x>1$ and $y>1$. Find the smallest possible value of
$$
\frac{\log _{x} 2011+\log _{y} 2011}{\log _{x y} 2011} .
$$
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
9. Answer: (A)
$$
\begin{aligned}
\frac{\log _{x} 2011+\log _{y} 2011}{\log _{x y} 2011} & =\left(\frac{\log 2011}{\log x}+\frac{\log 2011}{\log y}\right) \cdot\left(\frac{\log x y}{\log 2011}\right) \\
& =\left(\frac{1}{\log x}+\frac{1}{\log y}\right) \cdot(\log x+\log y) \\
& =2+\frac{\log x}{\log y}+\frac{\log y}{\log x} \\
& \geq 4(\text { using } A M \geq G M),
\end{aligned}
$$
and the equality is attained when $\log x=\log y$, or equivalently, $x=y$.