BMO 2017

Let $\forall (a,b,c)\in M$, $\alpha \le f(a,b,c)\le \beta$ and suppose that there are no better bounds, i.e. $\alpha$ is the largest possible and $\beta$ is the smallest possible. Now, $$\begin{array}{rl}\alpha \le f(a,b,c)\le \beta & \iff \alpha abc\le 4(ab+bc+ca)-1\le \beta abc\\ & \iff (\alpha -8)abc\le 4(ab+bc+ca)-8abc-1\le (\beta -8)abc\\ & \iff (\alpha -8)abc\le 1-2(a+b+c)+4(ab+bc+ca)-8abc\le (\beta -8)abc\\ & \iff (\alpha -8)abc\le (1-2a)(1-2b)(1-2c)\le (\beta -8)abc\end{array}$$ For $\alpha <8$, we have $$(1-2a)(1-2b)(1-2c)\ge 0>(\alpha -8)abc.$$ So $\alpha \ge 8$. But if we take $\epsilon >0$ small and $a=b=\frac{1}{4}+\epsilon$, $c=\frac{1}{2}-2\epsilon$, we'll have: $$(\alpha -8)(\frac{1}{4}+\epsilon )(\frac{1}{4}+\epsilon )(\frac{1}{2}-2\epsilon )\le (\frac{1}{2}-2\epsilon )(\frac{1}{2}-2\epsilon )4\epsilon$$ Taking $\epsilon \to 0^{+}$, we get $\alpha -8\le 0$. So $\alpha =8$ and it can never be achieved. For the right side, note that there is a triangle whose side-lengths are $a,b,c$. For this triangle, denote $p=\frac{1}{2}$ the half-perimeter, $S$ the area and $r,R$ respectively the radius of incircle, outcircle. Using the relations $R=\frac{abc}{4S}$ and $S=pr$, we will have: $$\begin{array}{rl}(1-2a)(1-2b)(1-2c)\le (\beta -8)abc& \iff (p-a)(p-b)(p-c)\le \frac{(\beta -8)abc}{8}\\ & \iff \frac{S^2}{p}\le \frac{(\beta -8)abc}{8}\\ & \iff 2\frac{S}{p}\le \frac{(\beta -8)abc}{4S}\\ & \iff \frac{R}{r}\ge 2(\beta -8{)}^{-1}\end{array}$$ Since the least value of $\frac{R}{r}$ is $2$ (this is a well-known classic inequality), and it is achievable at $a=b=c=\frac{1}{3}$, we must have $\beta =9$.

Answer: $\alpha =8$ not achievable and $\beta =9$ achievable.