Find the minimum value of $P=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)$

Question

For $a,b,c$ are positive numbers satisfying $a+b+c\leq \frac{3}{2}$, find the minimum value of $$P=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)$$

---

We have: $\frac{3}{2}\ge a+b+c \ge 3\sqrt[3]{abc}\Rightarrow abc\leq\frac{1}{8}$

We have $3+\frac{1}{a}+\frac{1}{b}\ge 7\sqrt[7]{\frac{1}{16a^2b^2}}$

$\Rightarrow P=\prod 7\sqrt[7]{\frac{1}{16a^2b^2}}=7^3$

When $a=b=c=\frac{1}{2}$

It looks wrong for $abc \le \frac{1}{8}$, and I need another way.

Answer 1

If you simply write out the inequality you obtained for the product explicitly, what you've got, rewriting $abc\le{1\over8}$ as ${1\over abc}\ge8$, is

$$P\ge\left(7\sqrt[7]{1\over16a^2b^2}\right)\left(7\sqrt[7]{1\over16b^2c^2}\right)\left(7\sqrt[7]{1\over16c^2a^2}\right) =7^3\sqrt[7]{1\over16^3(abc)^4}\ge7^3\sqrt[7]{8^4\over16^3}=7^3\sqrt[7]{2^{12}\over2^{12}}=7^3$$

with equality achieved when $a=b=c={1\over2}$.

Answer 2

By AM-GM $\frac{3}{2}\geq a+b+c\geq3\sqrt[3]{abc},$ which gives $\sqrt[3]{abc}\leq\frac{1}{2}$.

Thus, by Holder $$\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)\geq$$ $$\geq\left(3+\frac{1}{\sqrt[3]{abc}}+\frac{1}{\sqrt[3]{abc}}\right)^3\geq(3+2+2)^3=343.$$ The equality occurs for $a=b=c=\frac{1}{2}$.

Id est, the answer is $343$.