Q:

quadratic system of three equations with three unknowns

If $x,y,z$ are real and positive and they verify the system
$$\begin{cases}
x^2+xy+y^2=36\\
y^2+yz+z^2=64\\
z^2+zx+x^2=100
\end{cases}$$
Find the value of the sum $S=xy+yz+zx$
Of course I tried the obvious, to sum them all up, to bring each equation to the canonical form, to guess.  I can see that the system is symmetric, but I don't know how does it help.  Someone who already solved it told me that $S$ is proportional to $\sqrt{3}$. How to proceed? 

A:

If you are willing to use the Wolfram Cloud Sandbox then the answer comes out as
In[1]:= (x y + y z + z x)^2 /. Solve[{x^2 + x y + y^2 == 36,
  y^2 + y z + z^2 == 64, z^2 + z x + x^2 == 100}] // Simplify//Sqrt//InputForm
Out[1]//InputForm=
  32*Sqrt[3], 32*Sqrt[3], 32*Sqrt[3], 32*Sqrt[3]}

and so the answer is $\;32\sqrt{3}.$
Alternatively, define
 $\;q_1 := x^2+xy+y^2,\; q_2 := y^2+yz+z^2,\; q_3 := z^2+zx+x^2\;$ and also let
 $\;e_1 := x+y+z,\; e_2 := xy+yz+zx,\; e_3 := xyz.\;$ Given that
 $\;q_1=36, q_2=64, q_3=100,\;$ then
 $\;q_1+q_2+q_3 = 2e_1^2-3e_2 = 200\;$ and $\;q_1q_2+q_2q_3+q_3q_1 = e_1^4 + 3e_2^2 -3e_2e_1^2 = 12304.\;$ Now from the first equation $\;e_2=(2e_1^2-200)/3\;$ and when substituted into the second equation gives
$\;(40000-200e_1^2+e_1^4)/3 = 12304\;$ with solution $\;e_1^2 = 100+48\sqrt{3},\;$ and now $\;e_2 = 32\sqrt{3}.$

