We have $x^3-x^2 = y^3 - y^2 = z^3 - z^2 = xyz$. Just plot the graph of $f(x) = x^3-x^2$ and with easy maximal/minimal calculus see that $x,y,z \in \left(-\frac{1}{3},1\right)$, as the points are between $A$ and $B$ in the figure below, since in other case there's no way that a line parallel to $x$ axis touch $f(x)$ three times.

Goofy Solution without the use of Calculus: We see that $x^3 - x^2 = y^3 - y^2 = z^3 - z^2 = xyz = P$ so $x, y, z$ are the roots of the equation $t^3 - t^2 - P = 0$. By Vieta's, we get $x + y + z = 1$ and $xy + yz + zx = 0$. WLOG, assume $x \leq 0 \leq y, z$. Substituting $x = 1 - (y+z)$ into the second equation, we get $(y^2 - y) + yz + (z^2 - z) = 0$. The left hand side is minimized when $y = 0$, so thus, $0 \geq z^2 - z$ and $z \leq 1$. The same applies to $y$. However, when $z = 1$, $x+y = 0$ and $xy + x + y = 0$ so $x = y = 0$, which is a contradiction. Thus, all $x, y, z$ are strictly less than $1$. Now, we prove the minimum. Let $x' = -x$. The two given equations can be rewritten as $y + z = x'+1$ and $yz = x'(x'+1)$. Using AM-GM, we get $2\sqrt{x'(x'+1)} \leq x'+1$, which simplifies to $x' \leq \frac{1}{3}$. This cannot be attained since $y$ and $z$ are distinct. Thus, all $x, y, z$ are strictly greater than $-\frac{1}{3}$.

Let $x, y, z$ be three real distinct numbers such that $$\begin{cases} x^2+x=yz \\ y^2+y=zx \\ z^2+z=xy \end{cases}$$Show that $$-1 \leq x,y,z \leq \frac{1}{3} $$(YJ)

Some solutions by me and my friends: \begin{align} x^2 - x = yz \\ y^2 - y = zx \\ z^2 - z = xy \end{align} Solution 1 (by me): $(2) - (3): (y - z)(y + z - 1) = -x(y - z)$. By the statement of the problem, $y \neq z \Rightarrow y + z - 1 = -x \Leftrightarrow y + z = 1 - x$. This, together with $(1)$, implies that $y$ and $z$ are the roots of the equation $t^2 - (1 - x)t + (x^2 - x) = 0$. Taking the discriminant, we have $\Delta = (1 - x)^2 - 4(x^2 - x) = (1 - x)(1 - x + 4x) = (1 - x)(1 + 3x)$. Again, $y$ and $z$ are different reals $\Rightarrow \Delta > 0 \Rightarrow -\frac{1}{3} < x < 1$. Similarly, $-\frac{1}{3} < y < 1$ and $-\frac{1}{3} < z < 1$. Solution 2 (by @gastivy): Like in the first solution, we prove that $x + y + z = 1$. Assume, FTSOC, WLOG, $x \geq 1 \Rightarrow y + z \leq 0$. $(2) + (3): y^2 + z^2 = (x + 1)(y + z) \leq 0$ because $x + 1 > 0$ and $y + z \leq 0$. This implies $y = z = 0$, contradiction. Notice that, WLOG, $x = 0 \Rightarrow y = 0$ or $z = 0$ by $(1)$, contradiction. So $x, y, z \neq 0$. Assume, FTSOC, WLOG, $x \leq -\frac{1}{3} \Rightarrow y + z \geq \frac{4}{3}$. Thus, $(2)*(3)$ gives $(1 - y)(1 - z) = x^2$. As $x$ is negative, $\frac{2 - y - z}{2} \geq \sqrt{(1 - y)(1 - z)} = -x \Rightarrow y + z \leq 2x + 2 \leq \frac{4}{3} \Rightarrow y + z = \frac{4}{3}$ by AM-GM, but equality happens iff $1 - y = 1 - z$, contradiction. Solution 3 (by a math enthusiast): Again we have $x + y + z = 1$. Multiplying the three we get $$xyz(x-1)(y-1)(z-1) = (xyz)^2.$$If, one of them is $0$, WLOG $x$, we get $y= 0$ or $z = 0$ by the first equation. So they are all different than $0$, which means $$(x-1)(y-1)(z-1) = xyz$$$$\implies -xy - yz - zx + x + y + z - 1 = 0$$$$\implies xy + yz + zx = 0.$$ Then $$(x+y+z)^2 - 2xy - 2yz - 2zx = 1 = x^2 + y^2 + z^2.$$ So $|x| < 1$ (and the others). Since $y \neq z$, $(y-z)^2 > 0 \implies y^2 + z^2> 2yz$. $$(2) + (3): (x+1)(y+z) = y^2 + z^2 > 2yz = -2x(y + z).$$ If $y + z \leq 0$, $x\geq 1$ because $x + y + z = 1$, a contradiction. So, dividing the inequality by $y +z$ we get the desired result. Solution 4 (by @lambda5): $x,y,z \neq 0$ by solution 2. $(2)\cdot(3)$: $(y-1)(z-1)=x^2=x+yz \Rightarrow x+y+z=1$ $(4)$. $(1) + (2) + (3)$: $x^2+y^2+z^2-1=yz+xy+xz \Leftrightarrow 0=(x+y+z)^2-1=3yz+3xy+3xz$ So $xy+yz+xz=0$ and $x^2 +y^2+z^2=1$ $(5)$ By $(5)$ the upper bound is done. Notice that $x\leq-\frac{1}{3}$, we have by $(5)$ that $y^2+z^2 \leq\frac{8}{9}$ and by $(4)$ we have $y+z\geq\frac{4}{3} \Rightarrow z^2 + y^2 +2yz\geq\frac{16}{9} \Rightarrow 2yz\geq\frac{8}{9}\geq y^2+z^2 \Rightarrow (y-z)^2\leq0$, contradiction!

sqing wrote: Let $x, y, z$ be three real distinct numbers such that $$\begin{cases} x^2+x=yz \\ y^2+y=zx \\ z^2+z=xy \end{cases}$$Show that $$-1 \leq x,y,z \leq \frac{1}{3} $$(YJ) Let $a = -x, b = -y, c = -z$ $\begin{cases} a^2 - a = bc \\ b^2 - b = ca \\ c^2 - c = ab\end{cases}$ So $-\frac{1}{3}\le a,b,c \le 1$, or, $-1 \le x,y,z \le \frac{1}{3}$

assume contrary and do casework

In fact, the solution set over $(x,y,z)\in\mathbb{R}^3$ is $\{(1,0,0),(0,1,0),(0,0,1)\}\cup\left\{\left.\left(\frac{1}{3}+\frac{2}{3}\cos\theta,\frac{1}{3}+\frac{2}{3}\cos\left(\theta+\frac{2\pi}{3}\right),\frac{1}{3}+\frac{2}{3}\cos\left(\theta+\frac{4\pi}{3}\right)\right)\right|0\leq\theta<2\pi\right\}$. In the former part, $x,y,z$ is not all distinct, so they don't count. In the parametric part, they are not all distinct precisely when $\theta\in\{0,\frac{\pi}{3},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{5\pi}{3}\}$, so we don't allow these angles. The range of each of $x,y,z$ in this part is $\left[-\frac{1}{3},1\right]$, but since the endpoints of this range only happens for the unallowed $\theta$ values, when $x,y,z$ are distinct their range is $\left(-\frac{1}{3},1\right)$.

maybe correct