Let $s=|x+y+z|$. Note that \begin{align*} |x^3+y^3+z^3-3xyz|&=\frac{1}{2}s|(x-y)^2+(y-z)^2+(z-x)^2| \\ &\leqslant \frac{1}{2}s(|x-y|^2+|y-z|^2+|z-x|^2) \\ &= \frac{1}{2}s(3(|x|^2+|y|^2+|z|^2)-|x+y+z|^2) \\ &=\frac{1}{2}s(3-s^2) \\ &\leqslant 1. \end{align*}

The given condition implies that every column vector of the matrix \[A = \begin{bmatrix*} x & y & z\\ y & z & x\\ z & x & y \end{bmatrix*}\]has magnitude 1, so by Hadamard's inequality \[|x^3 + y^3 + z^3 - 3xyz| = \left|\det A\right| \le 1.\]

liekkas wrote: Given complex numbers $x,y,z$, with $|x|^2+|y|^2+|z|^2=1$. Prove that: $$|x^3+y^3+z^3-3xyz| \le 1$$ For all real numbers $ x,y,z$, \[ |x^3 + y^3 + z^3 - 3xyz|\leq (x^2 + y^2 + z^2)^{3/2}.

Hadamard Inequality should be proved before using it？？？

Even in TST

Using Hadamard Inequality，the value of RHS is 3^(3/2)?

We shall use the fact: \[|x-y|^2+|y-z|^2+|z-x|^2+|x+y+z|^2=3(|x|^2+|y|^2+|z|^2)\]which follows by assuming $x=a+bi,y=c+di,z=e+fi$ and then we get the left hand side is \begin{align*} &\left((a-c)^2+(c-e)^2+(e-a)^2+(a+c+e)^2\right)+\\ &\left((b-d)^2+(d-f)^2+(f-d)^2+(b+d+f)^2\right)\\ &=3(a^2+b^2+c^2+d^2+e^2+f^2) \end{align*}which is precisely the RHS. Thus, we get: \begin{align*} |x^3+y^3+z^3-3xyz|&=\frac 12|x+y+z|\left(|(x-y)^2+(x-z)^2+(y-z)^2|\right)\\ &\leq\frac 12|x+y+z|\left(|x-y|^2+|x-z|^2+|y-z|^2\right)\\ &=\frac 12|x+y+z|\left(3|x|^2+3|y|^2+3|z|^2-|x+y+z|^2\right)\\ &=\frac 12|x+y+z|\left(3-|x+y+z|^2\right)\\ \end{align*}To finish, we note that \[|x+y+z|\leq |x|+|y|+|z|\leq \sqrt{3(|x|^2+|y|^2+|z|^2)}=\sqrt 3\]which thus shows us $|x+y+z|\in (0,\sqrt 3]$. Now, we can use the AM-GM Inequality in the form \[|x+y+z|\cdot\left(\frac{3-|x+y+z|^2}2\right)\leq\left(\frac{3+2|x+y+z|-|x+y+z|^2}4\right)^2\]Factoring the last expression, we get \[\frac{(1+|x+y+z|)(3-|x+y+z|)}4=1-\frac{(|x+y+z|-1)^2}4\leq 1\]completing the proof.

Solved with goodbear, Th3Numb3rThr33. Let \(\omega=e^{2\pi i/3}\) be a primitive third root of unity. Lemma: If \(|x|^2+|y|^2+|z|^2=1\), then \[\frac{\left\lvert x+y+z\right\rvert^2+\left\lvert x+y\omega+z\omega^2\right\rvert^2+\left\lvert x+y\omega^2+z\omega\right\rvert^2}3=1.\] Proof. This follows from direct expansion. Here is a slick solution by Espen Slettnes using the probabilistic method. Here, \(\operatorname{Cov}(X,Y)=\mathbb E\big[(X-E[X])\overline{(Y-E[Y])}\big]\). Observe that \[\mathop{\mathbb E}_{\substack{|\alpha|=1\\ \beta^3=1}}\left(x\alpha+y\alpha\beta+z\alpha\beta^2\right)=0.\]Meanwhile, \begin{align*} \operatorname{Var}\left(x\alpha+y\alpha\beta+z\alpha\beta^2\right)&=\Big[\operatorname{Var}(x\alpha)+\operatorname{Var}(y\alpha\beta)+\operatorname{Var}\left(z\alpha\beta^2\right)\Big]\\ &\qquad+\Big[2\operatorname{Re}\operatorname{Cov}\left(x\alpha,y\alpha\beta\right)+2\operatorname{Re}\operatorname{Cov}\left(y\alpha\beta,z\alpha\beta^2\right)\\ &\qquad+2\operatorname{Re}\operatorname{Cov}\left(z\alpha\beta^2,x\alpha\right)\Big]\\ &=1+0. \end{align*}Recall that \(\operatorname{Var}(X)=\mathbb E\left[\left\lvert X^2\right\rvert\right]-\left\lvert\mathbb E[X]\right\rvert^2\), so \[\mathop{\mathbb E}_\beta\left[\left\lvert x+y\beta+z\beta^2\right\rvert^2\right]=\mathop{\mathbb E}_{\alpha,\beta}\left[\left\lvert x\alpha+y\alpha\beta+z\alpha\beta^2\right\rvert^2\right]=1,\]as needed. \(\blacksquare\) Remark: See this article for more details. Finally, the desired inequality is just AM-GM: \[\left\lvert x^3+y^3+z^3-3xyz\right\rvert^2=\left\lvert x+y+z\right\rvert^2\cdot\left\lvert x+y\omega+z\omega^2\right\rvert^2\cdot\left\lvert x+y\omega^2+z\omega\right\rvert^2\le1.\]

Let $S = \sum (\overline{x} y + x \overline{y}) = |x + y + z|^2 - 1$. Now \begin{align*}S + 1 = |x + y + z|^2 \leq (|x| + |y| + |z|)^2 \leq 3(|x|^2 + |y|^2 + |z|^2) = 3\end{align*}which shows that $S \leq 2$. Hence, \begin{align*}1 &\geq \frac 14 (S + 1)(2 - S)^2 \\ &= \frac 14|x + y + z|^2 \left(2|x|^2 + 2|y|^2 + 2|z|^2 - \sum (\overline{x} y + x \overline{y})\right)^2 \\ &= \frac 14 |x + y + z|^2 \left(\sum (x \overline{x} - y \overline{x} - x \overline{y} + y \overline{y})\right)^2 \\ &= \frac 14 |x + y + z|^2 \left(|x - y|^2 + |y - z|^2 + |z - x|^2\right)^2. \end{align*}Since this quantity is nonnegative, we may take square roots of both sides, which yields \begin{align*} 1 &\geq \frac 12 |x + y + z| \left(|x - y|^2 + |y - z|^2 + |z - x|^2\right) \\ &\geq \frac 12 |x + y + z| \cdot \left|(x - y)^2 + (y - z)^2 + (z - x)^2\right| \\ &= |x^3 + y^3 + z^3 - 3xyz|\end{align*}as desired.