# Is the geometric mean bounded above by this value?

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 $$\sum_k \alpha_k = 1$$.

• Your first inequality does not hold: take $M=1$, $\alpha_1=2$, $x_1 = 3$. Mar 23, 2022 at 11:39
• @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. Mar 23, 2022 at 11:42

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.