Deceptively simple inequality involving expectations of products of functions of just one variable For a proof to go through in a paper I am writing, I need to prove, as an auxiliary step, the following deceptively simple inequality:
$$E(X^a) E(X^{a+1} \ln X) > E(X^{a+1})E(X^a \ln X) $$
where $X>e$ has a continuous distribution and $0<a<1$. The intuition, in one sentence, is that if you start from 
$$E(X^a) E(X^a \ln X) = E(X^a)E(X^a \ln X) $$
it "pays more" (in terms of expected values) to place the added $X$ multiplying larger quantities $(X^a \ln X)$ than smaller quantities $(X^a)$. Simulations have confirmed the intuition, at least up to now. However, although I have tried to prove this inequality for days, using other well-known inequalities as well as relationships between expectations of products, products of expectations, and covariances, I have not been successful so far.
Something that seems related is that we know that 
$$E\left(\prod_i^n f_i(X)\right)>\prod_i^nE(f_i(X)) $$
as long as the functions $f_1\ldots f_n$ are continuous monotonic functions of $X$, and are all, for instance, increasing and satisfy $f_i(X)>0$ (e.g, John Gurland's "Inequalities of Expectations of Random Variables Derived by Monotonicity or Convexity", The American Statistician, April 1968). The inequality I am trying to prove is, in a sense, "in between" the two sides in the inequality above.
Any suggestion would be very greatly appreciated.
 A: Thanks for the insightful and fun problem. Here is a proof (I think) via the Cauchy-Schwarz 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
Cauchy-Schwarz 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 Cauchy-Schwarz 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 non-positive.  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 non-negative at nonzero $a+t$.
