Range of $x,y,z$ which is roots of cubic equation 
If $x+y+z=2$ and $xy+yz+zx=1$ and $x,y,z\in\mathbb{R}$ and $x\leq y\leq z$. Then complete set of values if $x,y,z$ are

Try: let $xyz=k$. Then we will form an cubic equation whose roots are $t=x,t=y,t=z$
$$f(t)=(t-x)(t-y)(t-z)= t^3-2t^2+t-k$$
$f'(t)= 3t^2-4t+1$ and $f''(t)=6t-4$
using $1$ st and $2$ nd derivative test, we will draw graph of $f(t)$
Put $f'(t)=0$ we have $3t^2-4t+1=(3t-1)(t-1)=0$ 
We have $\displaystyle t=1/3, t=1$
So $f''(1/3)=-2<0$ and $f''(1)=2>0$
So $t=1/3$ is a point of local maximum and $t=1$ is a point of local minimum.
Now i did not understand how to find range of $x,y,z$ . Thanks
 A: From $xy+(x+y)(2-x-y)=1$ we get:
$$y^2+y(x-2)+x^2-2x+1=0 $$
Since this quadratic equation on $y$ (and on $x$) has real solutions we have discriminat $D\geq 0$ so $$(x-2)^2-4(x-1)^2\geq 0$$
so $$3x^2-4x\leq 0\implies  x\in \Big[0,{4\over 3}\Big]$$
The same you can tell for $y,z$. Since $x$ is smallest we have $x\leq {2\over 3}$ so $$x\in \Big[0,{2\over 3}\Big]$$
Since $z$ is biggest we have $3z\geq 2$ so $$z\in\Big[{2\over 3},{4\over 3}\Big]$$ 
For $y$ we can say also this $$2=x+y+z\geq 0+y+y =2y$$ so $y\leq 1$ thus $$y\in [0,1]$$
A: I'd like to give another solution, using the work done by the original poster.
As they calculated (effectively, but not explicitely stated), the function $f$ has 3 intervals of monotony: strictly increasing in $(-\infty,\frac13]$, strictly decreasing in $[\frac13,1]$ and again strictly increasing in $[1,\infty)$. Also, $\lim_{t \to \infty} f(t) = \infty$ and $\lim_{t \to -\infty} f(t) = -\infty$.
In order for such a function to have 3 real roots, each interval of monotony needs to contain one root. That means we must have $f(\frac13) \ge 0$ (otherwise, $(-\infty,\frac13]$ contains no root), and $f(1) \le 0$ (same argument for $[1,\infty)$). Under these conditions, $[\frac13,1]$ contains the third (middle) root.
$f(\frac13) \ge 0$ is equivalent to $\frac13^3 - 2\frac13^2 + \frac13 - k \ge 0$, which means $k \le \frac4{27}$.
Similiarly, $f(1) \le 0$ means $1^3 - 2\times1^2 + 1 - k \le 0$, i.e. $k \ge 0$. 
This means the function $f$ has 3 real roots exactly when $0 \le k \le \frac4{27}$.
When looking at the graph of $f$ (or considering the monotony of $f$ in $(-\infty,\frac13]$), it becomes obvious that the smallest root ($x$) becomes as small as possible when $k$ is as small as possible, which means $k=0$. Solving $f(t)=0$ in that case leads to $x=0, y=z=1$. That means we have $x\ge 0$.
Similiarly, the smallest root $x$ becomes maximal when $k$ is maximal ($\frac4{27}$), in which case we get $x=y=\frac13, z=\frac43$.
So we know that $0 \le x \le \frac13$. If $k$ varies continously between $0$ and $\frac4{27}$, $x$ varies contiuously between the extreme, so we have that the range of x is the interval $[0,\frac13]$.
A similar argument shows that the range for $z$ is $[1, \frac43]$. When $k$ varies between $0$ and $\frac4{27}$, as shown above, the middle root  $y$ varies between $[\frac13,1]$.
A: by picture. We have $x^3 - 2x + x = x(x-1)^2 \; .$ The derivative is $3x^2 - 4 x + 1= (3x-1)(x-1).$ This had to factor as the original function had a double root at $1.$ The local max is $x = 1/3$ and function value $4/27.$
In order to have three real roots for
$$ x^3 - 2x+x = k \; , $$ 
allowing multiplicity, we need $0 \leq k \leq 4/27.$ By symmetry or calculation, the second point to achieve $4/27$ is $x=4/3.$
If we call your three real roots $a \leq b \leq c,$ we find
$$  0 \leq a \leq 1/3 , \; \; 1/3 \leq b \leq 1, \; \;  1 \leq c \leq 4/3    $$


