TITLE: An integral inequality (one variable)
QUESTION [16 upvotes]: Anyone has an idea to prove the following inequality?
Let $g:\left(0,1\right)\rightarrow\mathbb{R}$ be twice differentiable
and $r\in\left(0,1\right)$ such that 
$$
r\left(g"\left(x\right)+\dfrac{g'\left(x\right)}{x}\right)\geq\left(g'\left(x\right)\right)^{2},\forall x\in\left(0,1\right).
$$
Prove that 
$$
\left(\intop_{0}^{1}e^{-g\left(x\right)}xdx\right)\left(\intop_{0}^{1}e^{g\left(x\right)}xdx\right)\leq\dfrac{1}{4\left(1-r\right)},\quad\quad {(\star)}
$$
provided the LHS is finite. 

Further comment: This inequality comes from a contest for undergraduate students in my university. 
If we take the function 
$$
g\left(x\right)=-r\ln\left(-\ln\left(x\right)\right)
$$
 then we have 
$$
r\left(g"\left(x\right)+\dfrac{g'\left(x\right)}{x}\right)=\left(g'\left(x\right)\right)^{2},\forall x\in\left(0,1\right).
$$
By using Mathematica, we can see that for this function
$$
\left(\intop_{0}^{1}e^{-g\left(x\right)}xdx\right)\left(\intop_{0}^{1}e^{g\left(x\right)}xdx\right)=\dfrac{1}{4}\Gamma\left(1-r\right)\Gamma\left(1+r\right).
$$
Moreover, we also can check that 
$$
\dfrac{1}{4}\Gamma\left(1-r\right)\Gamma\left(1+r\right)\leq\dfrac{1}{4\left(1-r\right)},\forall r\in\left(0,1\right).
$$

REPLY [5 votes]: Let $f(x)=g(e^{-x})$, then
$$r\left(g''(x)+\frac{g'(x)}{x}\right)\ge g'^2(x)\iff rf''\ge f'^2$$
and
\begin{align}
I&=\left( \int_0^1 e^{-g(x)} xdx \right) \left( \int_0^1 e^{g(x)} xdx \right)\\
&=\left( \int_0^{\infty} e^{-f(t)} e^{-2t}dt \right) \left( \int_0^{\infty} e^{f(t)} e^{-2t}dt \right)\tag{$x=e^{-t}$}
\end{align}
Let $h(x)=e^{-f(x/2)/r}$, then
$$rf''\ge f'^2\iff h''\le0$$
and
\begin{align}
I&=\frac14\left( \int_0^{\infty} h(s)^r e^{-s}ds \right) \left( \int_0^{\infty} h(s)^{-r} e^{-s}ds \right)\tag{$s=2t$}\\
&=\frac14\int_0^{\infty}\int_0^{\infty} \left(\frac{h(t)}{h(s)}\right)^r e^{-t}e^{-s}dtds
\end{align}
Now the problem becomes

Let $h:\Bbb R_+\to\Bbb R_+$ be a twice-differentiable and concave function, let $r\in(0,1)$, prove that$$J=\int_0^{\infty}\int_0^{\infty} \left(\frac{h(t)}{h(s)}\right)^r e^{-t}e^{-s}dtds\le\frac1{1-r}$$

Let $s>0$, since $h$ is concave, we have
$$\forall t>0,\ h(t)\le h'(s)(t-s)+h(s)$$
As $t\to0$, since $h$ is positive, we have
$$h'(s)\le\frac{h(s)}{s}$$
Note that $h$ is positive and concave so is increasing, we have
\begin{align}
\forall t>0,\ \frac{h(t)}{h(s)}&\le\begin{cases}
1 & \text{if $t\le s$} \\
t/s  & \text{if $t>s$}
\end{cases}\\
&\le\frac{t}{s}+1
\end{align}
Therefore,
\begin{align}
J&\le\int_0^{\infty}\int_0^{\infty} (s+t)^r s^{-r}e^{-t}e^{-s}dtds\\
&=\int_0^{\infty}\Gamma(r+1,s) s^{-r}ds\tag{1}\\
&=\frac{1}{1-r}\tag{2}
\end{align}
$(1):$ Incomplete Gamma function,
$$\int_0^{\infty} (s+t)^r e^{-(s+t)}dt=\int_s^{\infty} u^r e^{-u}du\tag{$u=s+t$}$$
$(2):$ Integration by parts,
\begin{align}
\int_0^{\infty}\Gamma(r+1,s) s^{-r}ds&=\left[\frac{\Gamma(r+1,s) s^{1-r}}{1-r}\right]_0^{\infty}+\int_0^{\infty}s^re^{-s}\frac{s^{1-r}}{1-r}ds\\
&=0+\frac{\Gamma(2)}{1-r}
\end{align}

Update:
As mentioned by @Binjiu, we can give a sharper bound by using directly:
$$\forall s,t>0,\ \frac{h(t)}{h(s)}\le
\begin{cases}
1 & \text{if $t\le s$} \\
t/s  & \text{if $t>s$}
\end{cases}$$
This yields
\begin{align}
J&\le\int_0^{\infty}\int_0^{s} e^{-t}e^{-s}dtds+\int_0^{\infty}\int_s^{\infty} t^r s^{-r}e^{-t}e^{-s}dtds\\
&=\frac{1}{2}+\int_0^{\infty}\gamma(1-r,t)t^{r}e^{-t}dt\tag{3}\\
&=1+r\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1-r}\\
&=1+\frac{r}{2}\left(\psi\left(1-\frac r2\right)-\psi\left(\frac {1-r}2\right)\right)\tag{4}
\end{align}
$(3):$ Lower incomplete Gamma function,
$$\gamma (1-r,t)=\int_0^ts^{-r}e^{-s}ds=-r\gamma (-r,t)-t^{-r}e^{-t}$$
$$\gamma (-r,t)=t^{-r}\sum_{n=0}^{\infty}\frac{(-1)^{n}t^{n}}{n!(n-r)}$$
$(4):$ Digamma function,
$$\psi\left(\frac {z+1}2\right)-\psi\left(\frac {z}2\right)=2\sum_{n=0}^{\infty}\frac{(-1)^n}{n+z}$$
Below is the comparison of these two bounds. The maximum difference is $\ln2$ as $r\to1$.