Arithmetic mean greater than equal to harmonic mean inequality problem

Question

Given $a+b+c=1$ and $a>0,b>0,c>0$ we have to prove following

$$\left(1+\frac{1}{a}\right)\cdot\left(1+\frac{1}{b}\right)\cdot\left(1+\frac{1}{c}\right)\geq64$$

My attempt is as follows:

As $A.M\geq H.M$, so we can say

$$\frac{a+b+c}{3}\geq\frac{1}{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}}$$

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq9$$

Adding 3 on both sides

$$\left(1+\frac{1}{a}\right)+\left(1+\frac{1}{b}\right)+\left(1+\frac{1}{c}\right)\geq12$$

Dividing by 3 on both sides

$$\frac{\left(1+\frac{1}{a}\right)+\left(1+\frac{1}{b}\right)+\left(1+\frac{1}{c}\right)}{3}\geq4$$

Now can we say that harmonic mean of $\left(1+\frac{1}{a}\right),\left(1+\frac{1}{b}\right),\left(1+\frac{1}{c}\right)=4\quad?$ as initially we started with the condition $A.M\geq H.M.$

Answer 1

You showed that the arithmetic mean of the positive number $1+\frac1a$, $1+\frac1b$, $1+\frac1c$ is at least $4$. This does not by itself imply the desired result that the geometric mean of these numbers is also at least $4$.

Answer 2

It is $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\geq 64$$ equivalent to

$$\frac{ac+bc+ab}{abc}+\frac{2}{abc}\geq 63$$ by AM-GM we get $$\frac{ac+bc+ab}{abc}\geq 3\sqrt[3]{\frac{1}{abc}}$$ and we have to show that $$3\sqrt[3]{\frac{1}{abc}}+\frac{2}{abc}\geq 63$$ substituting $$t=abc$$ we get $$3\sqrt[3]{\frac{1}{t}}\geq 63-\frac{2}{t}$$ and this is equivalent to $$-{\frac { \left( 27\,t-1 \right) \left( 9261\,{t}^{2}-540\,t+8 \right) }{{t}^{3}}} \geq 0$$ this is true, since we get $$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$ or $$abc=t\le \frac{1}{27}$$