any one ?

I guess AM-GM

Lamp909 wrote: I guess AM-GM I guess you wrong!

Let $a,b,c$ be the roots to $t^3 - 3t - 1 = 0.$ which are clearly irrational. By Vieta we have $a+b+c = 0$, $ab+bc+ca = -3$, and $abc = 1$. We can calculate $a^3+b^3+c^3 = 3abc + (a+b+c)(a^2+b^2+c^2 - ab - bc - ca) = 3$ $\sum_{cyc} a^2b^2 = (ab+bc+ca)^2 - abc(a+b+c) = 9.$ $\sum_{cyc} a^3b^3 = 3a^2b^2c^2 + (ab+bc+ca)(\sum_{cyc} (ab)^2 - ab^2c) = -24.$ $\sum_{cyc} a^4bc = abc(a^3+b^3+c^3) = 3.$ Therefore, $(a^2b+b^2c+c^2a)(a^2c+b^2a+c^2b) = \sum_{cyc} a^3b^3+\sum_{cyc} a^4bc+3a^2b^2c^2 = -18.$ Since $(a^2b+b^2c+c^2a)+(a^2c+b^2a+c^2b) = (a+b+c)(a^2+b^2+c^2) - (a^3+b^3+c^3) = -3$, it follows that $\{a^2b+b^2c+c^2a, a^2c+b^2a+c^2b\} = \{-6,3\}$. WLOG arrange $a,b,c$ in such a way that $a^2b+b^2c+c^2a = -6$. The key part is that $$\prod_{cyc} (ax+by+cz) = abc\sum_{cyc} x^3 + \sum_{cyc}a^2b\sum_{cyc}x^2y + \sum_{cyc}a^2c\sum_{cyc}x^2z + (3abc+\sum_{cyc}a^3)xyz $$$$ = \sum_{cyc}x^3 -6\sum_{cyc}x^2y + 3\sum_{cyc}x^2z + 6xyz $$$$= (x+y+z)^3 - 9(x^2y+y^2z+z^2x).$$ Now let $x,y,z \in \mathbb Q$ satisfies $(x+y+z)^3 - 9(x^2y+y^2z+z^2x) = 0$. Then, by the factorization above, WLOG we have $ax+by+cz = 0$. Since $a+b+c = 0$, this implies $ax + by = az + bz$, so $a(x-z) = b(z-y)$. If $x,y,z$ are not all equal then it follows that $a/b$ is a rational number. Since $c = -a-b$, all of $c/a$ and $c/b$ are also rational numbers. Therefore, $c^3 = (c/a)(c/b)(abc) \in\mathbb Q$, so $3c+1 \in\mathbb Q$, so $c\in\mathbb Q$, which is a contradiction. Hence we must have $x=y=z$. $\blacksquare$

Suppose that $x=\min\{x,y,z\}$, let $y=x+a$ and $z=x+b$, where $a,b \in \mathbb{Q}_{\ge 0}$. $$(x+y+z)^3=9(x^2y+y^2z+z^2x) \Longleftrightarrow a^3-6a^2b+3ab^2+b^3=0 $$If either $a$ or $b$ is $0$, then the other one must be as well, if neither of them is $0$ , wlog assume that $a,b \in \mathbb{Z}_{>0}$ and that $\gcd(a,b)=1$ then $a|b$ and $b|a$, thus we have that $a^3=0\implies a=b=0$, contradiction.

Assume the solutions are integers. You can conclude that $x\equiv y\equiv z \pmod 3$ and then notice the same equation holds for smaller integer tuple $(\frac{x-r}3,\frac{y-r}3,\frac{z-r}3)$, where $r$ is the residue $\pmod 3$. Alternatively you can immediately replace $r$ with $x$ to obtain $(0,\frac{y-x}3,\frac{z-x}3)$ yielding the above solution.