Question: Can we give an example to show that the measurability of the set $f^{-1}(\{a\})$ for every a does not imply that $f$ is a measurable function?

Answer: 
Unfortunately the AM-GM inequality is the best you can do. If your data is $\{x,\frac{1}{x}\}$ the geometric mean will be $1$, yet you can make your arithmetic mean any value in $[1,+\infty)$ by choosing $x$ large enough.
Since the geometric mean for both $(2,2)$ and $(1,4)$ is $2$, while the arithmetic means are $2$ and $2.5$, the answer is a clear no. The only thing you can say is that the geometric mean is smaller or equal to the arithmetic.