It is clear that the geometric mean is bounded above by the arithmetic mean:
$$ \prod_{k=1}^{M} x_k^{\alpha_k} \leq \sum_{k=1}^{M}\alpha_k x_k $$
Moreover, it is clear that the arithmetic mean is bounded below by its maximal term:
$$ \max_k \alpha_k x_k \leq \sum_{k=1}^{M}\alpha_k x_k $$
So, my question is where does this bound lie in the first inequality? Specifically under what conditions is it true that:
$$ \prod_{k=1}^{M} x_k^{\alpha_k} \leq \max_k \alpha_k x_k $$
EDIT: I should have added the required constraint on the weights: $0 < \alpha_k <1$ and $\sum_k \alpha_k = 1$.
For $M=2$, $\alpha_1=\alpha_2 = \frac{1}{2}$ and $x_1=x_2=2$, we have $\displaystyle \max_{k}\alpha_k x_k = 1 < 2 = \prod_{k} x_k^{\alpha_k}$. But if $x_1=x_2 = 1$, there is equality. I doubt that there is much more to tell.