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 $\frac{4}{3}$.

Consider the polynomial:
$$ p(t)=(t-x)(t-y)(t-z) = t^3-2t^2+t+k = t(t-1)^2+k. $$
We know that $p(t)$ has three real roots, hence $-k=xyz$ is bounded between the two values of $t(t-1)^2$ in its stationary points. A stationary point obviously occurs for $t=1$, the other one occurs for $t=1/3$. The three reals roots of $p(t)$ hence belong to the interval $[0,u]$, where $u>1$ is the only real number such that
$$ u(u-1)^2 = \frac{1}{3}\left(\frac{1}{3}-1\right)^2 = \frac{4}{27}. $$
Now it is straightforward to check that $u=\frac{4}{3}$.
For a visual proof:
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