
A4

Let $M=\left\{(a, b, c) \in \mathbb{R}^{3}: 0<a, b, c<\frac{1}{2}\right.$ with $\left.a+b+c=1\right\}$ and $f: M \rightarrow \mathbb{R}$ given as

$$
f(a, b, c)=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{a b c}
$$

Find the best (real) bounds $\alpha$ and $\beta$ such that

$$
f(M)=\{f(a, b, c):(a, b, c) \in M\} \subseteq[\alpha, \beta]
$$

and determine whether any of them is achievable.


## Solution

Let $\forall(a, b, c) \in M, \alpha \leq f(a, b, c) \leq \beta$ and supose that there are no better bounds, i.e. $\alpha$ is the largest possible and $\beta$ is the smallest possible. Now,

$$
\begin{aligned}
\alpha \leq f(a, b, c) \leq \beta & \Leftrightarrow \alpha a b c \leq 4(a b+b c+c a)-1 \leq \beta a b c \\
& \Leftrightarrow(\alpha-8) a b c \leq 4(a b+b c+c a)-8 a b c-1 \leq(\beta-8) a b c \\
& \Leftrightarrow(\alpha-8) a b c \leq 1-2(a+b+c)+4(a b+b c+c a)-8 a b c \leq(\beta-8) a b c \\
& \Leftrightarrow(\alpha-8) a b c \leq(1-2 a)(1-2 b)(1-2 c) \leq(\beta-8) a b c
\end{aligned}
$$

For $\alpha(\alpha-8) a b c
$$

So $\alpha \geq 8$. But if we take $\varepsilon>0$ small and $a=b=\frac{1}{4}+\varepsilon, c=\frac{1}{2}-2 \varepsilon$, we'll have:

$$
(\alpha-8)\left(\frac{1}{4}+\varepsilon\right)\left(\frac{1}{4}+\varepsilon\right)\left(\frac{1}{2}-2 \varepsilon\right) \leq\left(\frac{1}{2}-2 \varepsilon\right)\left(\frac{1}{2}-2 \varepsilon\right) 4 \varepsilon
$$

Taking $\varepsilon \rightarrow 0^{+}$, we get $\alpha-8 \leq 0$. 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 $R=\frac{a b c}{4 S}$ and $S=p r$, we will have:

$$
\begin{aligned}
(1-2 a)(1-2 b)(1-2 c) \leq(\beta-8) a b c & \Leftrightarrow(p-a)(p-b)(p-c) \leq \frac{(\beta-8) a b c}{8} \\
& \Leftrightarrow \frac{S^{2}}{p} \leq \frac{(\beta-8) a b c}{8} \\
& \Leftrightarrow 2 \frac{S}{p} \leq \frac{(\beta-8) a b c}{4 S} \\
& \Leftrightarrow \frac{R}{r} \geq 2(\beta-8)^{-1}
\end{aligned}
$$

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.

