<s>[INST] 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.
 [/INST] 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.
</s>