<s>[INST] 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.
 [/INST] My Solution:: Given $x+y = 4-z$ and $x^2+y^2=6-z^2$. 
Now using the Cauchy-Schwarz inequality, we get $(x^2+y^2)\cdot (1^2+1^2)\geq (x+y)^2$
So we get $(12-2z^2)\geq (4-z)^2\Rightarrow 12-2z^2\geq 16+z^2-8z\Rightarrow 3z^2-8z+4\leq 0$
So $\displaystyle (3z^2-6z)-(2z-4)\leq 0\Rightarrow 3z(z-2)-2(z-2)\leq 0\Rightarrow \frac{2}{3}\leq z \leq 2$
Now $x+y+z=4$ and $x^2+y^2+z^2=6$ are symmetrical expressions on $x,y,z$
So we get $\displaystyle \frac{2}{3}\leq x,y,z\leq 2\Rightarrow x,y,z \in \left[\frac{2}{3}\;,2\right]$
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