Minimum of $ay+az+bz+bx+cx+cy$ with $ab+bc+ca=xy+yz+zx=1$ Let $a,b,c,x,y,z\in\mathbb{R}^+$, and $ab+bc+ca=xy+yz+zx=1$. What is the minimum value of $ay+az+bz+bx+cx+cy$?
When $a=b=c=x=y=z=\dfrac{1}{\sqrt{3}}$, the desired value is $2$.
When $a=b=x=y\rightarrow 1$ and $c=z\rightarrow 0$, the desired value also approaches $2$, so it seems likely that $2$ is the minimum.
 A: For any $a,b,c,x,y,z$, we have
$$ (a+b)(x+y)(ay+az+bz+bx+cx+cy) \\
= (xy+yz+zx)(a+b)^2 + (ab+bc+ca)(x+y)^2 + (bx-ay)^2 $$
So, in our situation with $xy+yz+zx = ab+bc+ca = 1$,
\begin{align*}
(a+b)(x+y)(ay+az+bz+bx+cx+cy)
&= (a+b)^2 + (x+y)^2 + (bx-ay)^2 \\
&\ge (a+b)^2 + (x+y)^2 \\
&\ge 2(a+b)(x+y)
\end{align*}
by AM/GM.  Since our numbers are all  positive, dividing by $(a+b)(x+y)$ yields the desired inequality.  The equality case is that $a+b=x+y$ and $ay=bx$, which is equivalent to $a=x$ and $b=y$ (in which case also $c=z$).
I'm not too happy with the symmetry breaking in this argument, but there it is.
A: This isn't exactly a proof, but I ran MATLAB with your function to minimize, adding large multiples of the absolute values $|ab + bc + ca - 1|$ and $|xy + yx + zx -1|$ so that minimizing the function preferred satisfying the constraints and THEN minimizing the main objective function. After optimizing with fminsearch in MATLAB using multiple starting points, the value of the primary objective function was always very, very close to 2 or higher. So your conjecture that the minimum is 2 is almost certainly correct.
