<s>[INST] 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 $\frac{-(x^2+y^2+z^2)}{4}$
 [/INST] 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.
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