Proof that $E[X^k]/E[X^{k-1}]\ge x_1 \Pr[X=x_1]^{1/k}$ where $x_1 = \max(X)$.

Question

This seems related to the Lehmer mean. An alternative form is

$$ \frac{\sum_i x^k_i p_i}{\sum_i x^{k-1}_i p_i} \ge x_1p_1^{1/k} $$

Where $x_1\ge x_2\ge\dots\ge 0$ and $\sum_i p_i=1$.

I tried applying Hölder, but that only gave me

$$ \frac{\sum_i x^k_i p_i}{\sum_i x^{k-1}_i p_i} \ge \left(\sum_i x^{k-1}_i p_i\right)^{1/(k-1)} \ge x_1 p_1^{1/(k-1)} $$

which seems to be a lot worse, when $x_1$ is much larger than the other $x_i$. E.g. when $k=2$, the rhs of the Hölder bound is $x_1p_1 << x_1\sqrt{p_1}$.

Answer 1

Somehow typing it in TeX made it easy to realize, that all I need is the power means inequality: $$ \left(\sum x_i^k p_i\right)^{1/k} \ge \left(\sum x_i^{k-1} p_i\right)^{1/(k-1)} $$ which implies $$ \left(\sum x_i^k p_i\right)^{k} \ge \left(\sum x_i^{k-1} p_i\right)^{k} \left(\sum x_i^k p_i\right) $$ and so $$ \frac{\sum x_i^k p_i}{\sum x_i^{k-1} p_i} \ge \left(\sum x_i^k p_i\right)^{1/k} \ge x_1 p_1^{1/k}. $$