复数不等式

jhsinutopia

$x,y,z\in\mathbb{C}$, $|x|^2+|y|^2+|z|^2=1$
Prove that: $|x^3+y^3+z^3-3xyz|\le1$

hbghlyj

The determinant of a circulant matrix $C$ with first row $(c_0, c_1, \dots, c_{n-1})$ is given by the product of the eigenvalues:
$$\det(C) = \prod_{k=0}^{n-1} (c_0 + c_1\omega^k + c_2\omega^{2k} + \dots + c_{n-1}\omega^{(n-1)k})$$
where $\omega = e^{2\pi i/n}$ is a primitive $n$-th root of unity.
For $n=2$, apply the Hadamard inequality to $C$,$$|(x+y)(x-y)| \le|x|^2+|y|^2\quad\forall x,y\inC$$
For $n=3$, apply the Hadamard inequality to $C$,$$|x^3+y^3+z^3-3xyz| \le (|x|^2+|y|^2+|z|^2)^{3/2}\quad\forall x,y,z\inC$$
For $n=4$, apply the Hadamard inequality to $C$,$$|(x+y+z+w)(x-y+z-w)((x-z)^2 + (y-w)^2)| \le (|x|^2+|y|^2+|z|^2+|w|^2)^2\quad\forall x,y,z,w\inC$$

kuing

以前证过实数范围内的（《撸题集》P.897），没想到在复数范围也成立😃

能否利用实数范围内的结果来证明呢？
（看来不太行，毕竟 `\abs{x^3+y^3+z^3-3xyz}\leqslant\abs x^3+\abs y^3+\abs z^3-3\abs{xyz}` 是不成立的😥

hbghlyj

Let $\omega = e^{i2\pi/3},A_0 = x+y+z,A_1 = x+\omega y + \omega^2 z,A_2 = x+\omega^2 y + \omega z$
So, $x^3+y^3+z^3-3xyz = A_0 A_1 A_2$.
$|A_0|^2 = (x+y+z)(\bar{x}+\bar{y}+\bar{z})$
$= |x|^2+|y|^2+|z|^2 + (\omega^2 x\bar{y} + \omega \bar{x}y) + (\omega x\bar{z} + \omega^2 \bar{x}z) + (\omega^2 y\bar{z} + \omega \bar{y}z)$
$|A_1|^2 = (x+\omega y + \omega^2 z)(\bar{x}+\bar{\omega} \bar{y} + \bar{\omega}^2 \bar{z})$
$= |x|^2 + |\omega y|^2 + |\omega^2 z|^2 + (x\omega^2\bar{y} + \bar{x}\omega y) + (x\omega\bar{z} + \bar{x}\omega^2 z) + (\omega y \omega \bar{z} + \bar{\omega}\bar{y}\bar{\omega}^2 z)$
$|A_2|^2 = (x+\omega^2 y + \omega z)(\bar{x}+\bar{\omega}^2 \bar{y} + \bar{\omega} \bar{z})$
$= |x|^2+|y|^2+|z|^2 + (\omega x\bar{y} + \omega^2 \bar{x}y) + (\omega^2 x\bar{z} + \omega \bar{x}z) + (\omega y\bar{z} + \omega^2 \bar{y}z)$
Summing $|A_0|^2, |A_1|^2, |A_2|^2$:
$$|A_0|^2 + |A_1|^2 + |A_2|^2 = 3(|x|^2+|y|^2+|z|^2)$$
We are given that $|x|^2+|y|^2+|z|^2=1$. Therefore, $|A_0|^2 + |A_1|^2 + |A_2|^2 =3$.
Let $P_0 = |A_0|^2$, $P_1 = |A_1|^2$, $P_2 = |A_2|^2$. These are non-negative real numbers.
We have $P_0+P_1+P_2 = 3$.
By AM-GM inequality:
$$|x^3+y^3+z^3-3xyz|^2=P_0 P_1 P_2\le(\frac{P_0+P_1+P_2}{3})^3=1$$