Let $A>0$ be fixed and consider $X_1,\ldots$ i.i.d. nonnegative random variables such that $E[1/X_1]<\infty$.
Is is true that $$\sup_{a\in \big (0,\frac A{\sqrt n} \big]} \sum_{i=1}^n 1_{X_i>a} \frac{a^3}{X_i^2}$$ converges in probability to $0$ ?
With the crude bound $\sum_{i=1}^n 1_{X_i>a} \frac{a^3}{X_i^2} \leq \frac {A^2}n \sum_{i=1}^n \frac{1}{X_i}$, the supremum is clearly $O_P(1)$. In my research I need it to be $o_P(1)$, but I haven't been able to prove it.
As suggested by Anthony Quas, the supremum in question can be rewritten as $$s_n:=\sup_{b\ge\sqrt n/A}S_n(b),$$ where $$S_n(b):=\frac1{b^3}\sum_{i=1}^n Z_i^2\,1(Z_i<b)$$ and $Z_i:=1/X_i$, so that the $Z_i$'s are iid positive random variables with $EZ_i<\infty$.
Take now any real $c>0$. Then for all large enough $n$ $$S_n(b)=T_n(b,c)+U_n(b,c),$$ where $$T_n(b,c):=\frac1{b^3}\sum_{i=1}^n Z_i^2\,1(Z_i<c),$$ $$U_n(b,c):=\frac1{b^3}\sum_{i=1}^n Z_i^2\,1(c\le Z_i<b).$$ Next, for all large enough $n$, uniformly in $b\ge\sqrt n/A$, $$0\le T_n(b,c)\le\frac{nc}{b^3}\frac1n\sum_{i=1}^n Z_i \sim\frac{nc}{b^3}\,EZ_1\to0$$ and $$0\le U_n(b,c)\le\frac n{b^2}\frac1n\sum_{i=1}^n Z_i\,1(c\le Z_i)\sim\frac{n}{b^2}\,EZ_1\,1(c\le Z_1) =O(EZ_1\,1(c\le Z_1))\underset{c\to\infty}\longrightarrow0$$ almost surely (a.s.), by the strong law of large numbers.
So, $s_n\to0$ a.s. and hence, indeed, in probability.
