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let $z_{1},z_{2},z_{3},z_{4},z_{5}$ are complex numbers,and such $$|z_{1}|^2+|z_{2}|^2+|z_{3}|^2+|z_{4}|^2+|z_{5}|^2=5$$

$$A=\begin{bmatrix}1&1&1&1&1\\z_1&z_2&z_3&z_4&z_5\\z_1^2&z_2^2&z_3^2&z_4^2&z_5^2\\z_1^3&z_2^3&z_3^3&z_4^3&z_5^3\\z_1^4&z_2^4&z_3^4&z_4^4&z_5^4\\\end{bmatrix}$$ show that $$\det{(AA^{*})}\le 5^5$$

my idea: we must prove $$|z_{1}-z_{2}|^2|z_{1}-z_{3}|^2|z_{1}-z_{4}|^2|z_{1}-z_{5}|^2|z_{2}-z_{3}|^2|z_{2}-z_{4}|^2|z_{2}-z_{5}|^2|z_{3}-z_{4}|^2|z_{3}-z_{5}|^2|z_{4}-z_{5}|^2\le 5^5$$ then let $$LHS=f(z_{1},z_{2},z_{3},z_{4},z_{5})$$ maybe can use this indentity $$\sum_{i=1}^{n}|z_{i}-z|^2=n\sum_{i=1}^{n}|z_{i}|^2-|\sum_{i=1}^{n}z_{i}|^2$$ where $$z=\dfrac{z_{1}+z_{2}+\cdots+z_{n}}{n}$$ But following I can't

math110
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    With such symmetry in the expression on the LHS and in the constraint, I can usually convince myself that extremal values happen either with the most symmetric input ($z_i=\exp(2\pi ik/5$) or the least symmetric input ($z_1=\sqrt{5}, z_{\geq2}=0$). The former here gives $5^5$ and the latter $0$. Of course this is proof of nothing, but it strongly suggests to me that this is probably true and it shows that the bound can be attained. – 2'5 9'2 Dec 03 '14 at 08:44
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    Schur http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN266833020_0001&DMDID=DMDLOG_0040&LOGID=LOG_0040&PHYSID=PHYS_0399 noted that if $z_1,\ldots,z_n$ are in the unit circle the maximum is indeed attained for the roots of $z^n-1=0$. It's not the same region here but it supports the suggestion of @alex.jordan. – Andrew Dec 17 '14 at 21:12

3 Answers3

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This turned out not to be an answer, but it may have helpful elements. Your LHS is the determinant of the big matrix below, which may help.

Let $A$ be the Vandermonde matrix $$\begin{bmatrix}1&1&1&1&1\\z_1&z_2&z_3&z_4&z_5\\z_1^2&z_2^2&z_3^2&z_4^2&z_5^2\\z_1^3&z_2^3&z_3^3&z_4^3&z_5^3\\z_1^4&z_2^4&z_3^4&z_4^4&z_5^4\\\end{bmatrix}$$ and $A^*$ be its conjugate transpose. Then using the Vandermonde determinant formula, your LHS is $\det AA^*$. Consider $$AA^*=\begin{bmatrix}5&\sum \bar{z_i}&\sum \bar{z_i}^2&\sum \bar{z_i}^3&\sum \bar{z_i}^4\\ \sum z_i&\sum z_i\bar{z}_i&\sum z_i\bar{z}_i^2&\sum z_i\bar{z}_i^3&\sum z_i\bar{z}_i^4\\ \sum z_i^2&\sum z_i^2\bar{z}_i&\sum z_i^2\bar{z}_i^2&\sum z_i^2\bar{z}_i^3&\sum z_i^2\bar{z}_i^4\\ \sum z_i^3&\sum z_i^3\bar{z}_i&\sum z_i^3\bar{z}_i^2&\sum z_i^3\bar{z}_i^3&\sum z_i^3\bar{z}_i^4\\ \sum z_i^4&\sum z_i^4\bar{z}_i&\sum z_i^4\bar{z}_i^2&\sum z_i^4\bar{z}_i^3&\sum z_i^4\bar{z}_i^4\\ \end{bmatrix}$$

A Hermitian matrix such as this is (unitarily) diagonalizable with real eigenvalues. So it is equivalent to some $B=\Delta(\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5)$ with $\lambda_i\in\mathbb{R}$. We have: $$\begin{align} \det(AA^*)&=\det(B)\\ &=\prod \lambda_i\\ &\leq\left(\frac{\sum\lambda_i}{5}\right)^5\qquad\text{AM-GM inequality}\\ &=\left(\frac{\operatorname{trace} B}{5}\right)^5\\ &=\left(\frac{\operatorname{trace} AA^*}{5}\right)^5\\ &=\left(\frac{5+\sum z_i\bar{z_i}+\sum z_i^2\bar{z_i}^2+\sum z_i^3\bar{z_i}^3+\sum z_i^4\bar{z_i}^4}{5}\right)^5\\ &=\left(\frac{5+5+\sum z_i^2\bar{z_i}^2+\sum z_i^3\bar{z_i}^3+\sum z_i^4\bar{z_i}^4}{5}\right)^5\\ \end{align}$$ But this is no good. In general these last three terms can reach as high as $25+125+625$, and that makes this bound too weak. They can only reach so high if one of the $z_i$ has norm $\sqrt{5}$ and the others are all $0$, and this is when there would be the largest discrepancy in the AM-GM comparison. So the step where we used AM-GM is giving too much away.

2'5 9'2
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Not the exact solution, but maybe a way to go: From the AM-GM inequality we have that $$ \prod_{i<j}|z_i-z_j|^2\leq\frac{1}{5^5}\left(\sum_{i<j}|z_i-z_j|^2\right)^5. $$ By the parallelogram identity we have $|z-w|^2=2(|z|^2+|w|^2)-|z+w|^2\leq2|z|^2+2|w|^2$ for each $z,w\in\mathbb C$. Thus, $$ \prod_{i<j}|z_i-z_j|^2\leq\frac{2^5}{5^5}\left(\sum_{i<j}(|z_i|^2+|z_j|^2)\right)^5=\frac{2^5\cdot 4^5}{5^5}\left(\sum_{i}|z_i|^2\right)^5=8^5. $$ Maybe one can use this to refine the argument a little...

sranthrop
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The required inequality is a straightforward corollary of Hadamard’s inequality.

Alex Ravsky
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