Minimize $xy$ over $x^2+y^2+z^2=7$ and $xy+xz+yz=4$. 
Let $x,$ $y,$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 7$ and $xy + xz + yz = 4.$ Find the smallest possible value of $xy.$


I used Cauchy to get $$(x^2+y^2+z^2)(1^2+1^2+1^2)\geq (x+y+z)^2$$ and $$(x^2+y^2+z^2)(y^2+z^2+x^2)\geq(xy+xz+yz)^2,$$ but this doesn't do much as both of the inequalities already work. :( Any guidance would be appreciated!!
Thanks in advance!!!
P.S. I've been given that this can be solved by Lagrange multipliers, but I haven't learned it yet. Hopefully, someone can give me hints on a method that doesn't involve Lagrange multipliers. :)
 A: I would start with noticing that
\begin{align*}
(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + xz + yz) = 15
\end{align*}
Therefore $x + y + z = \pm\sqrt{15} =: k$. Better saying, $z = k - x - y$.
Then we arrive at the relation
\begin{align*}
xy + xz + yz & = xy + z(x + y)\\\\
& = xy + (k - (x + y))(x + y)\\\\
& = xy + k(x + y) - (x + y)^{2} = 4\\\\
& \Rightarrow xy = (x + y)^{2} - k(x + y) + 4 
\end{align*}
Let us make the change of variable $t = x + y$.
Then we are interested in the minimum value of the RHS above, which is given by $1/4$ as follows:
\begin{align*}
t^{2} - kt + 4 & = \left(t^{2} - kt + \frac{k^{2}}{4}\right) + 4 - \frac{k^{2}}{4}\\\\
& = \left(t - \frac{k}{2}\right)^{2} + 4 - \frac{15}{4} \geq \frac{1}{4}
\end{align*}
Hence the minimum value of $xy$ satisfying the proposed constraints equals $1/4$.
Hopefully this helps !
EDIT
As an answer to the (well posed) comment of @RobPratt, we must show such value is attained.
Indeed, this is the case as the following system of equations show:
\begin{align*}
\begin{cases}
xy = \dfrac{1}{4}\\\\
x + y = \dfrac{\sqrt{15}}{2}
\end{cases} & \Longleftrightarrow
\begin{cases}
x = \dfrac{\sqrt{15} \mp \sqrt{11}}{4}\\\\
y = \dfrac{\sqrt{15} \pm \sqrt{11}}{4}
\end{cases}
\end{align*}
as well as the following system of equations:
\begin{align*}
\begin{cases}
xy = \dfrac{1}{4}\\\\
x + y = -\dfrac{\sqrt{15}}{2}
\end{cases} & \Longleftrightarrow
\begin{cases}
x = -\dfrac{\sqrt{15} \mp \sqrt{11}}{4}\\\\
y = -\dfrac{\sqrt{15} \pm \sqrt{11}}{4}
\end{cases}
\end{align*}
A: $x^2+y^2+z^2 = 7$ and $xy +xz + yz = 4$
implies
$x+y+z = \pm\sqrt{15}$
Now using the second equality that is given in the question, we get
$xy  =4 -(x+y)z = 4 -  (\pm\sqrt{15} - z)z$

*

*Case 1: When $x+y+z=\sqrt{15}$, minimising the right hand side with respect to $z$ gives $z = \frac{\sqrt{15}}{2}$. So, minimum value of $xy$ is $4 -  (\sqrt{15} - z)z= 4-\frac{15}{4}=\frac{1}{4}$.

*Case 2: When $x+y+z=-\sqrt{15}$, minimising the right hand side with respect to $z$ gives $z = -\frac{\sqrt{15}}{2}$. So, minimum value of $xy$ is $4 -  (-\sqrt{15} - z)z= 4-\frac{15}{4}=\frac{1}{4}$.

So the minimum value of $xy$ is $\frac{1}{4}$.
