Bounding a ratio that depends on a discrete probability distribution

Question

Suppose $p=(p_1,p_2,...,p_N)$ is a discrete probability distribution. Are there any assumptions we could make about $p$, say $1/(C_1 N) \leq p_i \leq C_2/N$ for every $i$ and $C_1, C_2>1$, that would enable us to bound from above the following ratio $$\frac{\sum_{i=1}^N p_i \big|\frac{1}{p_i}-N \big|^3}{\big(\sum_{i=1}^N \frac{1}{p_i} -N^2\big)^{3/2}}$$ by a quantity that neither depends on $N$ nor on $p$ ? The problem with separately bounding the nominator from above and the denominator from below is that the denominator equals zero when $p$ is the uniform distribution and I cannot afford an assumption about $p$ that excludes $p$ being close to uniform.

Answer 1

It cannot be done. Consider $Np_i = 1 + \frac{\epsilon_i}{\sqrt N}$, where the $\epsilon_i$ are small, and $\sum\epsilon_i = 0$. Then $\frac1{Np_i} \approx 1 - \frac{\epsilon_i}{\sqrt N} + \frac{\epsilon_i^2}N$. Then $$ \left(\sum\frac1{p_i} - N^2\right)^{3/2} \approx \left(\sum \epsilon_i^2\right)^{3/2} $$ Also $$ \sum p_i |\tfrac1{p_i}-N|^3 \approx \sum \frac1N |\sqrt N\epsilon_i|^3 = \sum \sqrt N |\epsilon_i|^3 .$$ So if we choose $\epsilon_1 = -\epsilon_2$, and $\epsilon_i = 0$ for $i \ge 3$, we see that the upper bound is larger than a constant times $\sqrt N$.

However, an upper bound of order $\sqrt N$ can be proven. \begin{align} \sum p_i |\tfrac1{p_i}-N|^3 &= \sum p_i^{-1/2} |\tfrac1{\sqrt{p_i}}-N\sqrt{p_i}|^3 \\&\le \left(\sum p_i^{-1/3} |\tfrac1{\sqrt{p_i}}-N\sqrt{p_i}|^2\right)^{3/2} \\&\le \sqrt{C_1 N} \left(\sum |\tfrac1{\sqrt{p_i}}-N\sqrt{p_i}|^2\right)^{3/2},\end{align} and $$ \sum \left(\frac1{\sqrt{p_i}} - N\sqrt{p_i} \right)^2 = \sum\frac1{p_i} - N^2.$$