I am posting here an answer to my question provided by @hbp in the related site Mathematics, because I think people here may find it quite interesting (I definitely did). I am copying below @hbp's response verbatim, i.e., I have contributed absolutely nothing to this (for some reason, when I tried to use blockquote it did not work well). @hbp's answer is here.
Response by @hbp starts here
Thanks for the insightful and fun problem. Here is a proof (I think) via the CauchySchwarz inequality. Consider the function
$$
f(t) \equiv \frac{ \mathbb E[X^{a+t} \ln X] } { \mathbb E[X^{a+t}] }.
$$
So the target inequality is $f(1) > f(0)$. We can show this by proving $f(t)$ is increasing, or $f'(t) \ge 0$.
But this is easy, because
$$
\begin{aligned}
f'(t)
&=
\frac{d}{dt} \left(
\frac{ \mathbb E[e^{(a+t)\ln X} \ln X] } { \mathbb E[e^{(a+t) \ln X}] }
\right)
\\
&=
\frac{ \mathbb E\left[ \frac{d}{dt} e^{(a+t)\ln X} \ln X \right] }
{ \mathbb E\left[e^{(a+t) \ln X} \right] }

\mathbb E[ e^{(a+t)\ln X} \ln X ]
\frac{ \mathbb E\left[ \frac{d}{dt} e^{(a+t) \ln X} \right] }
{ \mathbb E[e^{(a+t) \ln X}]^2 }
\\
%&=
%\frac{ \mathbb E\left[ e^{(a+t)\ln X} (\ln X)^2 \right] }
%{ \mathbb E\left[e^{(a+t) \ln X} \right] }
%
%\mathbb E\left[ e^{(a+t) \ln X} \ln X \right]
%\frac{ \mathbb E\left[ e^{(a+t) \ln X} \ln X \right] }
%{ \mathbb E\left[e^{(a+t) \ln X}\right]^2 }
\\
&=\frac{
\mathbb E[X^{a+t} (\ln X)^2] \, \mathbb E[X^{a+t}]

\mathbb E[X^{a+t} (\ln X)]^2
} {
\mathbb E\left[X^{a+t}\right]^2
} \ge 0.
\qquad (1)
\end{aligned}
$$
The numerator of (1) is nonnegative by the
CauchySchwarz inequality.
That is, with $U = X^{\frac{a+t}{2}} \ln X, V = X^{\frac{a+t}{2}}$, we have
$$
\mathbb E\left[U^2 \right] \mathbb E\left[V^2\right] \ge \mathbb E[U \, V]^2.
\qquad (2)
$$
It remains to argue that the equality cannot hold for all $t \in [0,1]$, which is easy.
Alternative to the CauchySchwarz inequality (2)
Alternatively, we can show (1) directly by observing that
$$
\mathbb E\left[X^{a+t}(y  \ln X)^2 \right] \ge 0,
$$
holds for all $y$ (for the quantity of averaging is nonnegative), i.e., the quadratic polynomial
$$
\begin{aligned}
p(y)
&=
\mathbb E\left[X^{a+t}\right] y^2
 2 \, \mathbb E\left[X^{a+t} \ln X\right] y
+ \mathbb E\left[X^{a+t} (\ln X)^2\right]
\\
&\equiv A \,y^2  2 \, B \, y + C,
\end{aligned}
$$
has no zero.
Thus the discriminant of $p(y)$, which is $4B^2  4AC$, must be nonpositive. This means $AC \ge B^2$, or
$$
\mathbb E\left[X^{a+t}\right] \,
\mathbb E\left[X^{a+t} (\ln X)^2\right]
\ge
\mathbb E\left[X^{a+t} \ln X\right]^2.
$$
Further discussion
There is a more intuitive interpretation of (1). We define the characteristic function of $\ln X$ as
$$
F(t) \equiv \log \left\{ \mathbb E\left[ X^{a+t} \right] \right\}.
$$
We find $f(t) = F'(t)$, and $f'(t) = F''(t) \ge 0$. In other words, (1) is a generalized statement of that the second cumulant of $\ln X$ is nonnegative at nonzero $a+t$.