Proof of an Inequality

Given a sequence $(b_n)_{n=1}^p$ of positive numbers such that $b_1>b_2>\cdots>b_p>0$, define $$\beta=\bigg(p!\frac{p}{p-1}\bigg)^{\frac{1}{\min\limits_{1\leq k\leq p-1}{(b_k-b_{k+1})}}}.$$ Suppose $a_i\in\mathbb{C},i=1,2,\cdots,p$ satisfies $|a_i|>\beta|a_{i+1}|, i=1,2,\cdots p-1$ and $|a_p|>1$. Prove the following matrix $$D= \begin{pmatrix} a_{1}^{b_1} & a_{1} ^{b_2}& \dots & a_{1} ^{b_p}& \\ a_{2} ^{b_1}& a_{2} ^{b_2}& \dots &a_{2} ^{b_p}\\ \\ a_{p} ^{b_1}& a_{p} ^{b_2}& \dots & a_{p} ^{b_p} \end{pmatrix}$$ satisfies the following inequality: $$\frac{1}{p}\prod_{i=1}^{p}|a_i|^{b_i}<|\det(D)|<2\prod_{i=1}^{p}|a_i|^{b_i}.$$

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If Riemann can't solve it, what can we do? :) Sorry for the off topic, couldn't resist. –  N. S. Apr 9 '12 at 16:01
What is the context for this inequality? Why it should be true? –  Beni Bogosel Apr 9 '12 at 16:34
Writing \det with a backslash in $\TeX$ does not just prevent italicization; it also provides for proper spacing and is standard usage. Correct: $a\det A$. Incorrect: $a det A$. I editing the question accordingly. –  Michael Hardy Apr 9 '12 at 16:38
this question is from Graduate Admission Test for the AMSS OF CAS! www.amss.ac.cn –  Riemann Apr 9 '12 at 16:40