Obtaining the bound without using only the trivial inequality 
If $x,y,z$ are 3 real numbers such that $$x+y+z = 4$$ and $$x^2 + y^2 + z^2 = 6$$ show that each of $x,y,z$ lie in the closed interval $[\frac{2}{3},2]$

I obtained the upper bound, but I am unable to obtain the lower bound. 
My approach
$x^2 + y^2 + (4-(x+y))^2 = 6$
$\therefore 2x^2 + 2y^2 + 2xy - 8x - 8y = -10 \implies (2-y-x)^2 - 4 -xy = -5 \implies xy -1 \geq 0$ 
$$\therefore \color {blue}{xy \geq 1}$$
similarly we get $yz \geq 1$ and $zx \geq 1$
Now suppose that $z >2 \:\:$ Then $x + y < 2$ and $x^2 + y^2 <2 \implies (x+y)^2 - x^2 - y^2 = \color{red}{2xy < 2}$
Contradiction. A similar argument works for $x$ and $y$. Therefore we have $x,y,z \leq 2$

If possible, I'd like a similar argument for the lower bound, i.e, using only the trivial inequality. Thank you.

$y+z=4-x$ and $(4-x)^2-2yz=6-x^2$, which gives $yz=x^2-4x+5$.
Thus, $(4-x)^2-4(x^2-4x+5)\geq0$, which is $(x-2)(3x-2)\leq0$ and we are done!
