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.

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 $a+t$.

• Thanks much for the suggestion. It seems to me the CSI does not apply, as you do not have $E([X^{a+t}]^2)$ on the left. You can get $[E(X^{a+t})]^2$, which is smaller than $E([X^{a+t}]^2)$. – Sandokan Dec 23 '15 at 18:16
• I did not understand the second approach. Would you mind elaborating? – Sandokan Dec 23 '15 at 18:33
• @Sandokan Regarding your first question, he applies Cauchy-Schwarz to $X^{(a+t)/2} \ln X$ and $X^{(a+t)/2}$. – angryavian Dec 23 '15 at 19:52
• @Sandokan, I updated the answer. Hopefully it is clearer. Also, many thanks to angryavian for answering the first question for me. Happy holidays! – hbp Dec 23 '15 at 20:21
• hbp, this is beyond great, much appreciated. I had asked the same question at StackOverflow (as I did not know where, or whether, I would get a response), and received other good answers -- but I think yours provides the most direct approach. If you feel like it, maybe you could copy and paste your answer there as well, as I think those who answered will find it quite interesting. Thanks again. – Sandokan Dec 24 '15 at 2:11