Prove or disprove that the given expression is "always" positive I have previously asked a question and I tried to solve it by my own and it led to the question below:

Prove or disprove that
$$\small\int_{\mathbb{R}}l(y)^xf_0(y)\mathrm{d}y\int_{\mathbb{R}}f_0(y)l(y)^x\ln (l(y)^x)\ln (l(y))\mathrm{d}y-\int_{\mathbb{R}}l(y)^xf_0(y)\ln (l(y))\mathrm{d}y\int_{\mathbb{R}}f_0(y)l(y)^x\ln (l(y)^x)\mathrm{d}y$$
is greater than $0$.

Given:
$\rightarrow f_0$ and $f_1$ are some density functions
$\rightarrow l(y)=\frac{f_1(y)}{f_0(y)}$ is an increasing function.
$\rightarrow x\in(0,1)$
 A: This is an application of Cauchy-Schwarz in disguise. It suffices to find the appropriate Hilbert space. And then show that the relevant functions actually belong to it. The latter is the tedious part of this exercise. 
Consider the positive function $w(y)=\ell(y)^xf_0(y)=f_1(y)^xf_0(y)^{1-x}$. By Holder, it is integrable over $\mathbb{R}$. Now consider the weighted $L^2$ space $L^2_w(\mathbb{R})$ equipped with the inner product

$$
(g,h)_w:=\frac{\int_\mathbb{R}g(y)h(y)w(y)dy}{\int_\mathbb{R}w(y)dy}
$$

and the resulting norm $\|g\|_w=\sqrt{(g,g)_w}$. Note that $g$ is in $L^2_w$ if, by definition, $g^2w$ is integrable over $\mathbb{R}$.
Observing $\ln(\ell(y)^x=x\ln(\ell(y))$, what you want to show boils down to 
$$
\left(\int_\mathbb{R} \ln(\ell(y))w(y)dy\right)^2< \left(\int_\mathbb{R} (\ln(\ell(y)))^2w(y)dy\right) \left(\int_\mathbb{R} (w(y)dy\right).
$$
Dividing by $\left(\int_\mathbb{R} (w(y)dy\right)^2$, this reads

$$
(g,1)_w^2< \|g\|_w^2=\|g\|_w^2\|1\|_w^2\qquad\mbox{with}\;\;g(y)=\ln(\ell(y)).
$$

This follows from Cauchy-Schwarz. Note that this will be a strict inequality as $g$ and $1$ are linearly independent, since $\ell$ is increasing. 
It only remains to prove that $g$ belongs to $L^2_w$, i.e. $\int_\mathbb{R} g(y)^2w(y)dy<\infty$. This is where the assumption $\ell$ increasing comes into play. 
If $g$ is bounded, the claim is obvious. If not, since $\ell$ is increasing and positive, this means that either $\lim_{+\infty} \ell=+\infty$ or $\lim_{-\infty} \ell=0$. So assume, to begin with, that $\lim_{+\infty} \ell=+\infty$. Now write
$$
(\ln \ell(y))^2w(y)=(\ln \ell(y))^2\ell(y)^xf_0(y)   =\frac{(\ln \ell(y))^2}{\ell(y)^\frac{1-x}{2}}\ell(y)^\frac{1+x}{2}f_0(y)=\frac{(\ln \ell(y))^2}{\ell(y)^\frac{1-x}{2}}f_1(y)^\frac{1+x}{2}f_0(y)^\frac{1-x}{2}.
$$
By Holder, the function $f_1(y)^\frac{1+x}{2}f_0(y)^\frac{1-x}{2}$ is integrable. Now
$$
\lim_{+\infty} \frac{(\ln \ell(y))^2}{\ell(y)^\frac{1-x}{2}}=0.
$$
So $g(y)^2w(y)$ is integrable over $[0,+\infty)$ by comparison with $f_1(y)^\frac{1+x}{2}f_0(y)^\frac{1-x}{2}$. If $\lim_{-\infty}\ell>0$, then $g$ is bounded on $(-\infty,0]$ and integrability over $(-\infty,0]$ follows. If $\lim_{-\infty}\ell=0$, then write
$$
(\ln \ell(y))^2w(y)=(\ln \ell(y))^2\ell(y)^xf_0(y).
$$
Now
$$
\lim_{-\infty}(\ln \ell(y))^2\ell(y)^x=0
$$
hence integrability over $(-\infty,0]$ follows by comparison with $f_0$. n any case, we have proven that $g$ belongs to $L^2_w$, which completes the argument.
