<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] 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 $\frac{2}{3}$. 
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