If $x,y,z \in \Bbb{R}$ such that $x+y+z=4$ and $x^2+y^2+z^2=6$, then show that $x,y,z \in [2/3,2]$. If $x,y,z\in\mathbb{R}$ such that $$x+y+z=4,\quad x^2+y^2+z^2=6;$$then show that the each of $x,y,z$ lie in the closed interval $[2/3,2]$.
I have been able to solve using $2(y^2+z^2)\geq(y+z)^2$.
Is there any another method to solve it.

We have $(x-1)^2+(y-1)^2+(z-1)^2=x^2+y^2+z^2-2(x+y+z)+3=6-8+3=2$, i.e.,
$$(x-1)^2+(y-1)^2+(z-1)^2=1.$$
This implies $(x-1)^2\le1$ and $x\in[0,2]$. Similarly for the other two variables.
Similarly we get 
$$\left(x-\frac53\right)^2+\left(y-\frac53\right)^2+\left(z-\frac53\right)^2=1$$
which implies $(x-\frac53)^2\le1$ and $x\in[2/3,8/3]$.
