Cyclical System

x+y+z=1 $\Rightarrow \color{#D61F06}{\text{y+z=1-x....(1)}}$ $xy+yz+zx=-1\\ \Rightarrow x(y+z)+zx=-1 \\ \Rightarrow x(1-x)+yz=-1...(\small{\text{Using 1}})$ $\Rightarrow \color{#D61F06}{\text{yz}=x^2-x-1....(2)}$ Applying AM-GM, $(y+z)^2\geq 4yz$ and using (1) and (2) $\Rightarrow (1-x)^2 \geq 4(x^2-x-1)$ $\Rightarrow 3x^2-2x-5\leq 0$ $(x+1)(3x-5)\leq0$ $-1\leq x\leq \frac{5}{3}$ Hence maximum value of x is $\Large\boxed{\frac{5}{3}=1.67}$ Corresponding to x's maximum value, y=z=- $\frac{1}{3}$

Substituting $z=1-x-y$ gives the ellipse $x^2+y^2+xy-x-y-1=0$ .The discriminant when solving for $y$ is $-3x^2+2x+5=-(x+1)(3x-5)$ , so that the largest value is $x=\frac{5}{3}\approx\boxed{1.66}$

We have $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=1$

$\Rightarrow x^2+y^2+z^2=3$ So now we have $\begin{cases} x^2+y^2+z^2=3\\x+y+z=1\end{cases}$ Using Cauchy-Schwarz inequality we have $x^2+y^2+z^2\geq x^2+\frac{(y+z)^2}{2} = x^2+\frac{(1-x)^2}{2}$ $\Leftrightarrow 6\geq 3x^2-2x+1$ Solving the inequality and we get $-1\leq x\leq\frac{5}{3}$