<s>[INST] if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$ 
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$ 

things i have done: first thing to do is to show that $x,y,z$ are non-negative. $$xy+yz+zx=1 \Rightarrow zx=1-yz-xy \Rightarrow zx=1-y(z+x)\Rightarrow zx=1-y(2-y)=(y-1)^2$$
this equality shows that $x$ and $z$ have same sign(both of them are positive or negative).like this we can conclude that $xy=(x-1)^2$ and $yz=(x-1)^2$.So all variables are positive or negative.If all of them are negative then $x+y+z \neq 2$ so all of $x,y,z$ are non negative. I don't know what to do for showing  that none of $x,y,z$ will be be bigger than $\frac{4}{3}$.
 [/INST] You have
$$
1=xy+z(x+y)=xy+(x+y)(2-(x+y))=xy+x(2-x)+2y-y^2,
$$
so that
$$
0=y^2+(x-2)y+x^2-2x+1=\bigg(y+\frac{x-2}{2}\bigg)^2+\frac{3}{4}x^2-x
$$
and hence $x-\frac{3}{4}x^2 =-\bigg(y+\frac{x-2}{2}\bigg)^2 \leq 0$. So we have
$\frac{3x}{4}(\frac{4}{3}-x) \leq 0$, i.e. $x(x-\frac{4}{3}) \geq 0$, hence
$x\in [0,\frac{3}{4}]$. By symmetry, the same holds for $y$ or $z$.
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