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This question is based on the question that I asked here Need advice: what should be my next step? I did a little more progress and wanted to share with you. As this is a new question, without any details simply the problems turns out to be showing that

$$\small{D(y_u)=1-\frac{\sqrt{L(y_u)}\int_{-\infty}^{-y_u}f_1(y)\mbox{d}y+\frac{\sqrt{L(y_u)}}{\sqrt{L(y_u)}+1}\int_{-y_u}^{y_u}\left(\sqrt{f_0(y)}+\sqrt{f_1(y)}\right)\sqrt{f_1(y)}\mbox{d}y+\int_{y_u}^{\infty}f_1(y)\mbox{d}y}{\sqrt{L(y_u)\int_{-\infty}^{-y_u}f_1(y)\mbox{d}y+\frac{L(y_u)}{\left(\sqrt{L(y_u)}+1\right)^2}\int_{-y_u}^{y_u}\left(\sqrt{f_0(y)}+\sqrt{f_1(y)}\right)^2 \mbox{d}y +\int_{y_u}^{\infty}f_1(y)\mbox{d}y}}}$$

is an increasing function in $y_u$.

My progress:

Let $D(y_u)=1-T(y_u)$. Then if $D(y_u)$ is increasing, $T(y_u)$ should be decreasing. I use here Cauchy-Schwarz inequality for the denominator.

$$\sqrt{L(y_u)\int_{-\infty}^{-y_u}f_1(y)\mbox{d}y+\frac{L(y_u)}{\left(\sqrt{L(y_u)}+1\right)^2}\int_{-y_u}^{y_u}\left(\sqrt{f_0(y)}+\sqrt{f_1(y)}\right)^2 \mbox{d}y +\int_{y_u}^{\infty}f_1(y)\mbox{d}y}\leq \sqrt{L(y_u)\int_{-\infty}^{-y_u}f_1(y)\mbox{d}y}+\sqrt{\frac{L(y_u)}{\left(\sqrt{L(y_u)}+1\right)^2}\int_{-y_u}^{y_u}\left(\sqrt{f_0(y)}+\sqrt{f_1(y)}\right)^2}+\sqrt{\int_{y_u}^{\infty}f_1(y)\mbox{d}y}$$

Assume that the error resulting from the Cauchy-Schwarz approximation (with equality) yields an error $\Delta(y_u)\geq 0$, Then I can rewrite $D(y_u)$ as


The first and third terms in the denominator are greater than their corresponding terms in the nominator. The second term in the denominator seems to be also greater than the similar term in nominator. If this is the case I was thinking that the error term $\Delta(y_u)$ is less than (all other terms in denominator$-$ all terms in nominator). As a results the denominator dominates and the function could be shown to be decreasing.

Are my conclusions correct or in the right direction? what could be more done?

Thank you very much for your help.

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