How prove this inequality $\prod_{1\le i
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
 A: 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.
A: 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...
A: The required inequality is a straightforward corollary of Hadamard’s inequality.
