Let $X$ be a random variable that is almost surely positive. Let $\alpha \ge 1$. Is it true that $$\mathbb{E}(X^{\alpha+1} \ln X) \mathbb{E}(X^\alpha) \ge \mathbb{E}(X^{\alpha+1}) \mathbb{E}(X^\alpha \ln X)?$$
This problem arises in the following context. The following number measures the spread in the distribution of $X$: $\frac{\mathbb{E}(X^{\alpha+1})}{\mathbb{E}(X^\alpha)}$. I would like to prove that whenever $\alpha$ increases, the measure of spread also increases.
This is true if the inequality above is true, which follows from taking first order conditions w.r.t $\alpha$, in the measure of spread.
For $\alpha \ge 1$, if random variable $X\ge 0$ and $0<\mathsf{E}[X^{\alpha}]<\infty$, then the following inequality is true, \begin{equation*} \mathsf{E}[X^{\alpha+1}\ln X]\mathsf{E}[X^{\alpha}] \ge\mathsf{E}[X^{\alpha+1}]\mathsf{E}[X^{\alpha}\ln X]. \tag{1} \end{equation*} To prove (1), consider the new probability $\widetilde{\mathsf{P}} $ and expection $\widetilde{\mathsf{E}} $ as following: \begin{align*} \widetilde{\mathsf{P}}(A)&=\frac{\mathsf{E}[1_{A}X^{\alpha}]}{\mathsf{E}[X^{\alpha}]}, \\ \widetilde{\mathsf{E}}[Y]&=\frac{\mathsf{E}[YX^{\alpha}]}{\mathsf{E}[X^{\alpha}]}. \qquad(\text{if }\mathsf{E}[|Y|X^{\alpha}]<\infty). \tag{2} \end{align*} Using (2), (1) is equivalent to the following inequality, \begin{equation*} \widetilde{\mathsf{E}}[X\ln X] \ge \widetilde{\mathsf{E}}[X]\widetilde{\mathsf{E}}[\ln X].\tag{3} \end{equation*} Since $x,\ln x$ are increasing functions $\mathbb{R}_+ \to \mathbb{R}_+ $ both, from the inequality in this, (3) holds.