<s>[INST] 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
 [/INST] The required inequality is a straightforward corollary of Hadamard’s inequality.
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