Q:

What is the range of $y$ if $x+y+z=4$ and $xy+yz+xz=5$ for $x, y, z \in\mathbb{R}_+$

What is the range of $y$ if $x+y+z=4$ and $xy+yz+xz=5$ for $x, y, z \in\mathbb{R}_+$

How to explain the following method?
Let $x=z$ then: 
$$2x+y=4\quad;\quad 2xy+x^{2}=5$$
$$\implies \left( 4-y\right) y+\left( \dfrac {4-y} {2}\right) ^{2}=5 $$
Simplify: $(3y-2)(y-2)=0$. Then I get $\dfrac {2} {3}\leq y\leq 2$.
But why?
Thanks and sorry for my poor English

A:

$\bf{Another \; Way::}$
Given $x+y+z = 4\Rightarrow x+z=4-y$ and $xy+yz+zx = 5$
Using $$(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)\Rightarrow x^2+y^2+z^2 = 6$$
So we get $x^2+z^2 = 6-y^2$
Now Using $\bf{Weighted\; A.M\geq G.M\;,}$ We get
$$\frac{x^2+z^2}{2}\geq \left(\frac{x+z}{2}\right)^2$$
So $$\frac{6-y^2}{2}\geq \left(\frac{4-y}{2}\right)^2\Rightarrow (4-y)^2\leq 12-2y^2$$
So $$16+y^2-8y\leq 12-2y^2\Rightarrow 3y^2-8y+4\leq 0$$
So $$3y^2-6y-2y+4\leq 0\Rightarrow \frac{2}{3}\leq y\leq 2$$

