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. :)

$x^2+y^2+z^2 = 7$ 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 $\frac{1}{4}$.
