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 $\frac{-(x^2+y^2+z^2)}{4}$

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
