Trouble understanding proof of the inequality - $(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1) \ge 64 $, for $a,b,c > 0$ and $a+b+c = 1$

Question

I was looking into this problem in a book discussing inequalities, However I found the proof quite hard to understand.The problem is as follows:

Let $a,b,c$ be positive numbers with $a+b+c=1$, prove that

$$\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right) \ge 64$$

and the proof provided was the following:

Note that $$ abc \le (\frac{a+b+c}{3})^3 = \frac{1}{27} \tag{1}$$ by AM-GM inequality. Then

$$(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc} \tag{2}$$

$$\ge 1+\frac{3}{\sqrt[3]{abc}}+\frac{3}{\sqrt[3]{(abc)^2}} +\frac {1}{abc} \tag{3}$$

$$=(1+\frac{1}{\sqrt[3]{abc}})^3 \ge 4^3 \tag{4}$$

Steps 1 and 2 are easy for me to understand, but if someone could help me with steps 3 and 4 ,I would be very thankful.

Answer 1

For Step $(3)$,

Using AM-GM inequality

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 3 \sqrt[3]{\frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c}}=\sqrt[3]{\frac{1}{abc}}\tag{3.1}$$

$$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \geq 3 \sqrt[3]{\frac{1}{ab} \cdot \frac{1}{bc} \cdot \frac{1}{ca}}=\sqrt[3]{(abc)^2}\tag{3.2}$$

For Step $(4)$,

Setting $a=1$ and $b=\frac{1}{\sqrt[3]{abc}}$ and using $a^3+3a^2b+3ab^2+b^3=(a+b)^3$ we get,

$$1+\frac{3}{\sqrt[3]{abc}}+\frac{3}{\sqrt[3]{(abc)^2}}+\frac{1}{abc}=\left(1+\frac{1}{\sqrt[3]{abc}} \right)^3\tag{4.1}$$

Using AM-GM inequality

$$\frac{a+b+c}{3}=\frac{1}{3} \geq \sqrt[3]{abc} \Rightarrow \frac{1}{\sqrt[3]{abc}} \geq 3 \tag{4.2}$$ as $a,b,c>0$ and the result follows.

Answer 2

using for $$\frac{1}{a},\frac{1}{b},\frac{1}{c}$$ the AM-GM inequality we obtain $$\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq \sqrt[3]{\frac{1}{abc}}$$ and for $$\frac{1}{ab},\frac{1}{bc},\frac{1}{ca}$$ the same we get $(3)$

Answer 3

$$\frac{a + \frac 1 3 + \frac 1 3 + \frac 1 3}{4} \ge \sqrt[4]{a\cdot \frac 1 3 \cdot \frac 1 3 \cdot \frac 1 3} $$

and similarly for b and c. So $(a+1)(b+1)(c+1) \ge 64\sqrt[4]{\frac{abc}{27^3}}$. You must show this is $\ge 64 abc$, which follows from your inequality (1).

Answer 4

I have any solution.

Note that

$a+1=a+a+b+c\geq 4\sqrt[3]{a^2bc} \tag{1}$

by AM-GM

Analog we get that

$b+1=b+a+b+c\geq 4\sqrt[3]{ab^2c} \tag{2}$

and

$c+1=c+a+b+c\geq 4\sqrt[3]{abc^2} \tag{3}$

So

$$(a+1)(b+1)(c+1)\geq 64abc\Longleftrightarrow (\dfrac{a+1}{a})(\dfrac{b+1}{b})(\dfrac{c+1}{c})\geq 64$$

The equality will hold by $a=b=c=1$.

Proved