<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] Note we also have $x^2+y^2+z^2=2$ (expand $(x+y+z)^2=4$). 
Note $y^2+z^2 \geq 2yz=2-2x(y+z)=2-2x(2-x)$, so
$x^2+y^2+z^2 \geq 3x^2-4x+2$
But $3x^2-4x+2 >2$ for $x>\frac{4}{3}$.
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