Let $f\in L^2(0,\infty)$ be a positive, decreasing function. Is it then true that $$ \limsup_{x\to\infty} xf(x) = \limsup_{x\to\infty} \frac{1}{f(x)}\int_x^{\infty} f^2(t)\, dt $$ (and similarly for $\liminf$)?

This looks strange at first; for example, the quotient of the two quantities can easily become both large or small. However, from the context in which this question arose, we have reason to believe that the statement is actually true.