Suppose that we have $0<\tau<\infty$.
In addition, we are informed that the following equation for possible $\tau$ holds (there exists at least one $\tau$ such that the following equation holds): \begin{equation} \frac{f_1(\tau)}{f_0(\tau)}=\left[\frac{\int_{0}^{\tau}f_1(x)dx\int_{\tau}^{\infty}f_1(x)dx}{\int_{0}^{\tau}f_0(x)dx\int_{\tau}^{\infty}f_0(x)dx}\right]^{\frac{1}{2}} \qquad \qquad \qquad \qquad(1) \end{equation} where $f_1(x)$ and $f_0(x)$ are two unknown probability density functions, $x\in (0,\infty)$, and we have $$L(x)=\frac{f_1(x)}{f_0(x)}$$ is monotonically increasing w.r.t. $x$
Then, can we prove that there only exists one $\tau\in (0,\infty)$ such that (1) holds.
My approach:
Generally, for the concerned problem, we can study the monotonicity of the following function： \begin{equation} h(\tau)=\left[\frac{f_1(\tau)}{f_0(\tau)} \right]^2-\frac{\int_{0}^{\tau}f_1(x)dx\int_{\tau}^{\infty}f_1(x)dx}{\int_{0}^{\tau}f_0(x)dx\int_{\tau}^{\infty}f_0(x)dx} \end{equation} If $h^{\prime}(\tau)$ is always larger or smaller than $0$, then we deduce that there only exists one $\tau$ (recall that we already know that there at least exists one $\tau$ such that (1) holds).
However, from the $h^{\prime}(\tau)$, I can not judge the relationship between $h^{\prime}(\tau)$ and $0$.
Any helpful suggestions and answers will be greatly appreciated!
Not a solution, but here is a counterexample to the monotonicity of $h$
Using $f_1(x) = e^{-x}$ and $f_0(x) = 2e^{-2x}$, we have $f_1(x)/f_0(x) = e^{x}/2$, which is monotonically increasing. Next, $$ \begin{align} h(\tau)&=\left[\frac{f_1(\tau)}{f_0(\tau)} \right]^2-\frac{\int_{0}^{\tau}f_1(x)dx\int_{\tau}^{\infty}f_1(x)dx}{\int_{0}^{\tau}f_0(x)dx\int_{\tau}^{\infty}f_0(x)dx}\\ &=\frac{e^{-2\tau}}{4e^{-4\tau}} - \frac{(1-e^{-\tau})e^{-\tau}}{(1-e^{-2\tau})e^{-2\tau}}\\ &=\frac{e^{2\tau}}{4}-\frac{e^{2\tau}}{1+e^{\tau}} \end{align} $$ and so there is a solution to $h(\tau)=0$, namely $\tau=\ln(3)$. However, $h$ is not monotone, as $$h'(\tau) =\frac{1}{2}e^{2\tau}-\frac{(1+e^{\tau})2e^{2\tau}-e^{\tau}(e^{2\tau})}{(1+e^{\tau})^2}$$ So $h'(0) = -\frac{1}{4} < 0$ whereas $h'(\ln(2))=\frac{2}{9}>0$