Minimum value of $ab+bc+ca$ Given that $a,b,c \in \mathbb{R^+}$ and $(a+b)(b+c)(c+a)=1$

Find the Minimum value of $ab+bc+ca$

My try: Letting $x=a+b, y=b+c, z=c+a$ we get
$$a=\frac{x+z-y}{2}, b=\frac{x+y-z}{2},c=\frac{y+z-x}{2}$$
$$xyz=1$$
Now the problem is to minimize $$ab+bc+ca=\frac{xy+yz+zx}{2}-\frac{x^2+y^2+z^2}{4}$$
Now by $A.M-G.M$ we have
$$\frac{xy+yz+zx}{2}\geq \frac{3}{2}\times \sqrt[3]{x^2y^2z^2}=\frac{3}{2}$$
But I am stuck for $\frac{-(x^2+y^2+z^2)}{4}$
 A: This doesn't have a well-defined minimum. Solve
$$(a+b)(b+c)(c+a)=1$$
for $c$ to get
$$
c=\frac12\left(-(a+b)\pm\sqrt{(a+b)^2-4\left(ab-{1\over a+b}\right)}\right)
$$
Let $a=b=1/N$ where $N$ is some
"big enough" number, then $a+b\approx0$ and $c\approx\sqrt N$ and $ab+bc+ca\approx {1/\sqrt N}\approx 0$, up to factors of two. Obviously, it never reaches the minimum though, since $a,b,c>0$, so we don't have a well-defined minimum value.
A: With this separation you can't find a minimum ; since $x^2+y^2+z^2$ is not bounded. Just take the sequence $(x_n,y_n,z_n)=(n,n,1/n^2)$, $n\in\mathbb{N}^\ast$, so that $x_ny_nz_n=1$, then $x_n^2+y_n^2+z_n^2=2n^2+1/n^4$ is not upper bounded. You may maximize the 'whole' function $x^2+y^2+z^2-2(xy+yz+zx)$ by Lagrange multiplier (for example).
Lagrange multiplier
A: To find a potential boundary value, why not try $a=0$? - Then you are minimising $bc$ subject to $bc(b+c)=1$ and this is rather easier to work with.
This may give you an idea - now to prove it, given you can't have $a=0$ but can get as close as you wish.
Eliminating one of the variables in this way is one way of trying to understand what is going on when you have three variables like this. Sometimes the same idea can be used to reduce many variables to a special case with two - it can indicate, for example, whether you need to separate the values as much as possible or make the variables as close to each other as you can. And that can pint you in the direction of the right kind of technique (I always find it easy to mix up max/min $\gt/ \lt$ in this kind of question unless I'm very careful).
