If $x,y,z \in \Bbb{R}$ such that $x+y+z=4$ and $x^2+y^2+z^2=6$, then show that $x,y,z \in [2/3,2]$.

Question

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.

Answer 1

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