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 $\sum_k \alpha_k = 1$.

  • $\begingroup$ Your first inequality does not hold: take $M=1$, $\alpha_1=2$, $x_1 = 3$. $\endgroup$
    – Didier
    Mar 23, 2022 at 11:39
  • $\begingroup$ @Didier Sorry, just added the condition that we are taking a convex combination of the $x_k$'s. The first inequality is the weighted AM-GM inequality BTW. $\endgroup$
    – Jacob A
    Mar 23, 2022 at 11:42

1 Answer 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.

  • $\begingroup$ Yes this is a good counterexample, I've just found a few similar simple ones too. But I'm curious on the extra conditions under which it may hold (or the reverse inequality for that matter) $\endgroup$
    – Jacob A
    Mar 23, 2022 at 12:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .