Given the following set of equations, find: $xy + 2yz + 3zx$. 
The positive reals x, y, z satisfy the equations
$$x^2 + xy + \frac{y^2}{3}=25$$
$$\frac{y^2}{3}+z^2 = 9$$
$$z^2+ zx + x^2 = 16$$
Find $$xy + 2yz + 3zx$$


My understanding:


*

*What struck me first were the squares $9, 16, 25$. This is the “Egyptian triangle.” It is a hint to the theorem of Pythagoras, to geometry, and geometrical
interpretation.


*Instead of $x, y, z$ only $xy + 2yz + 3zx$ is required. This may be an area,
maybe even the area $6$ of the Egyptian triangle. It is also a hint that I
should not try to find $x, y, z$.


*$\frac{y^2}{3}$ occurs twice, so it may be helpful to set $a^2=\frac{y^2}{3}$.
The equations finally become:
$$x^2+ \sqrt{3}xt + t^2 = 25$$
$$t^2+z^2=9$$
$$z^2+ zx + x^2 = 16$$
Also, one more observation, $zx$ must be a perfect square.

I am not sure if I'm on the right track, but I don't know how to proceed further. Hints and help would be appreciated, thanks.

Adding the second and third equation and subtracting first equation , we get
$2z^2+xz=yx$
Substitute the value for $y$ in the first equation, we get
$$3x^4+7x^2z^2+3x^3z+4z^4+4z^3x=75x^2$$
which easily gets factored into
$$(x^2+z^2+xz)(3x^2+4z^2)=75x^2$$
$$(16)(3x^2+4z^2)=75x^2$$
One possible way to avoid all the algebraic mess is that on solving the above equation we get,
$z^2=\frac{27}{64}x^2$ you could take $a^2=\frac{27}{64}$. This simplifies the calculations a lot since $z=ax$ and $y=2a^2x+ax$
On subsitituting in second equation we get $27=4x^2(a^2+a^3+a^4)$
And we have to find $4x^2(a+a^2+a^3)=\frac{27}{a}$.
And we are done. Not the cleanest approach but it is a way if nothing else seems obvious.
