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

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...
