Suppose $X$ is some non-negative random variable with $\mathbb{E} X^2<\infty$ and that $F$ is its cumulative distribution function, with generalized inverse $F^{-1}$. I'm wondering how we might be able to bound the integral $\int_0^{1/2}(F^{-1}(1-x))^2dx.$
If we have a $k$-th moment for $k>2$, then I can use Markov's inequality to show that $$F^{-1}(1-x)\leq ||X||_k/x^{1/k}$$ and hence $$\int_0^{1/2}(F^{-1}(1-x))^2dx \leq ||X||_k^2 \int_0^{1/2} 1/x^{2/k}dx<\infty.$$ The problem with filling in $k=2$ is that the final integral in the bound is no longer finite.
How could I solve this problem (if it is even possible)?
I will assume that $F$ is continuous and strictly increasing to simplify things. If you need a result for more general $F$, you will need to figure that out on your own.
Using this trick for integrating inverse functions, you get $$ \int_{0}^{1/2}(F^{-1}(1-x))^2\,dx =\frac12 F^{-1}(1/2)^2+\int_{F^{-1}(1/2)^2}^\infty (1-F(x^{1/2}))\,dx $$ Using the substitution $x=u^2,dx=2u\,du$, this becomes $$ \frac12 F^{-1}(1/2)^2+\int_{F^{-1}(1/2)}^\infty 2u(1-F(u))\,du $$ Write $m=F^{-1}(1/2)$, which is the median of $X$. Using the same Fubini trick which proves $E[X^2]=\int_0^\infty 2tP(X>t)\,dt$, $$ \begin{align} \frac12 m^2+\int_{m}^\infty 2uP(X>u)\,du &=\frac12m^2+\int_m^\infty 2uE[{\bf 1}(X>u)]\,du \\&=\frac12m^2+E\left[\int_m^X2u\,du\right] \\&=\frac12m^2+E[(X^2-m^2)\cdot {\bf 1}(X>m)] \\&=E[X^2\cdot {\bf 1}(X> m)] \end{align} $$ This is an exact expression for your integral, which also comes with the simple bounds $$ E[X^2]-\frac12m^2\le E[X^2\cdot {\bf 1}(X> m)]\le E[X^2] $$