Minimum value of $ab+bc+ca-abc$

Question

If $a,b,c>0$ and $a+b+c=1$ find the minimum value of $$ab+bc+ca-abc$$ I just found that maximum value of $ab+bc+ca-abc$ is $8/27$ by Schur's inequality.And i couldn't find the minimum value.Any help would be appreciated.

Answer 1

From $a,b,c > 0$ and $a+b+c=1$ , it follows that $0 < a,b,c < 1$.

Let $f(a,b,c)=ab+bc+ca-abc$.

From $a+b+c=1$ , it follows that $f(a,b,c)=(1-a)(1-b)(1-c)$, hence $f(a,b,c) > 0$.

Letting $a$ approach $1$ from below, and $b=c=(1-a)/2$, we get that $f(a,b,c)$ approaches $0$ from above.

Thus $f$ has a greatest lower bound of $0$, but $f$ has no minimum.

Answer 2

For $b=c\rightarrow0^+$ we obtain a value $0$.

But, $$ab+ac+bc-abc=(a+b+c)(ab+ac+bc)-abc>0,$$ which says that $0$ is an infimum and the minimum does not exist.

Answer 3

This is equal to $$(1-a)(1-b)(a+b)$$ since it is the specialization $x=1$ of $(x-a) (x-b) (x-(1-a-b))$. This has no minimum value.