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.
 A: HINT:
So, we have $x^2+y^2+(4-x-y)^2=6$ 
$\implies 2y^2+2y(x-4)+2x^2-8x+10=0$ which is a quadratic equation of $y$
As $y$ is real, the discriminant must be $\ge0,$  this will give us the range of $x$
We need to use : if $(x-\alpha)(x-\beta)\le0$ where $\alpha\le \beta$ 
We shall have $\alpha\le x\le \beta$ (Would you try proving it?) 
Observe that the equations are symmetric with respect to $x,y,z$
So, the ranges of $x,y,z$ will be same
A: The mean of the numbers is $\frac{4}{3}$, while the mean of their squares is $2$,
that is
$$E(x)= \bar x =\frac{1}{n}\sum x= \frac{4}{3} \\
E(x^2) = \frac{1}{n}\sum x^2 = 2$$
($n$ is $3$)
 Therefore, the variance is 
$$\frac{1}{n}\sum(x - \bar x)^2  = E(x^2) - E(x)^2 = 2 - (\frac{4}{3})^2 = \frac{2}{9}$$
In other words, 
$$\frac{1}{3}\left( (x-\frac{4}{3})^2 + (y - \frac{4}{3})^2 + (z- \frac{4}{3})^2\right) = \frac{2}{9}$$
This is the average quadratic deviation from the mean. 
Rewrite it as
$$ (x-\frac{4}{3})^2 + (y - \frac{4}{3})^2 + (z- \frac{4}{3})^2 = \frac{2}{3}$$
With $a = x-\frac{4}{3}$, $b=y - \frac{4}{3}$, $c= z- \frac{4}{3}$, we have 
$$a+b+c=0\\
a^2 + b^2 + c^2 = \frac{2}{3}$$
Now the question is how large can be the coordinate of a point $(a,b,c)$ on the circle of radius $r=\sqrt{\frac{2}{3}}$ described by the above equations. It turns out that the largest value for a coordinate is $\sqrt{\frac{2}{3}} \cdot r=\frac{2}{3} $. Hence the numbers $x$, $y$, $z$ differ from $\frac{4}{3}$ by at most $\frac{2}{3}$. 
A: 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 $\displaystyle \frac{2}{3}\leq x,y,z\leq 2\Rightarrow x,y,z \in \left[\frac{2}{3}\;,2\right]$
A: 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 $x\in[2/3,8/3]$.
