Upper bound on ratio of probability sums Let us say I have some vector of probabilities $\vec{V}$ such that $\sum_i v_i = 1$.  I am trying to find a sharp upper bound on
$\frac{\sum_i v_i^3}{(\sum_i v_i^2)^2}$.  I can currently prove via Cauchy-Schwarz a very bad upper bound - this ratio must be less than the number of entries in the vector.  But I have numerical evidence that leads me to believe that something like 1.5 should be a sharp upper bound, not dependent on the vector length.
 A: The Henry's suspecting is right!
Indeed, let also $\sum\limits_{i=1}^nv_1^2=constant$ and $n>1$.
Thus, by the Vasc's E-V Method(see here https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf Corollary 1.5)
it's enough to find a maximal value for $v_1=v_2=...=v_{n-1}\leq v_n$.
From here, as Henry says, we have,  $v_n=x$, where $x\in[0,1]$,  $$v_1=...=v_{n-1}=\frac{1-x}{n-1}$$ and $$\frac{1-x}{n-1}\leq x,$$ which gives $$\frac{1}{n}\leq x\leq1.$$
Thus, we need to find $$\max_{\frac{1}{n}\leq x\leq1}\frac{\frac{(1-x)^3}{(n-1)^2}+x^3}{\left(\frac{(1-x)^2}{n-1}+x^2\right)^2}.$$
Now, $$\left(\frac{\frac{(1-x)^3}{(n-1)^2}+x^3}{\left(\frac{(1-x)^2}{n-1}+x^2\right)^2}\right)'=\frac{(1-nx)((n^2-2n)x^3+3nx^2-3nx+1)}{(nx^2-2x+1)^3}.$$
We can show that for $n>2$ $$x_{max}=\frac{2\sqrt{n-1}\cos\frac{\pi-\arccos\frac{2\sqrt{n-1}}{n}}{3}-1}{n-2}.$$
For $n=2$ $x_{\max}=\frac{1+\sqrt3}{2\sqrt3}.$
I hope it will help.
A: Empirically it looks to me as if there is an upper bound which is $O(\sqrt{n})$ and in particular your expression can be above $0.32 \sqrt{n}$ but will never exceed  $\frac{\sqrt{n}+2}{3}$ for all vector lengths $n$.
So for example with a vector of length $n=100$, this suggests the maximum possible is somewhere between $3.2$ and $4$.  Consider the case where $x_1=0.1585$ and  the other ninety-nine $x_i$ are all equal to $0.0085$; your expression would then be about $3.88$.
I suspect it would be possible to prove that the maximal case has

*

*one value is $x$ and the other $n-1$ values are $\frac{1-x}{n-1}$

*where $x$ is the largest root of $(n^2-2n)x^3+3nx^2-3nx+1=0$

Added later
Based on River Li and Michael Rozenberg's anaysis, we can tighten this: the maximum possible value is between $$\frac{3\sqrt 3}{16}\sqrt{n} +\frac58 \qquad\text{ and } \qquad\frac{3\sqrt 3}{16}\sqrt{n} +1-\frac{3\sqrt 3}{16}$$  so numerically between $$0.32475\sqrt{n} +0.625\qquad\text{ and } \qquad 0.32476\sqrt{n} +0.67524.$$ My empirical results were consistent with this.
