# An Inequality with Constraint XVIII

### Solution 1

The inequality is equivalent to $(a+1)(b+1)(c+1)\ge 64abc,\,$ which, given the constraint, is equivalent to

$\displaystyle\prod_{cycl}(2a+b+c)\ge 64abc.$

Using the AM-GM inequality,

\begin{align} 2a+b+c&\ge 4\sqrt[4]{a^2bc}\\ a+2b+c&\ge 4\sqrt[4]{ab^2c}\\ a+b+2c&\ge 4\sqrt[4]{abc^2}. \end{align}

The product of the three gives the required inequality.

### Solution 2

For $x\gt 0,\,$ $\displaystyle y=\log\left(1+\frac{1}{x}\right)\,$ is a convex function, hence, by Jensen's inequality,

$\displaystyle \sum_{cycl}\log\left(1+\frac{1}{a}\right)\ge 3\log\left(1+\frac{1}{\displaystyle\frac{a+b+c}{3}}\right)=3\log 4=\log 64.$

### Solution 3

By the AM-HM inequality,

$\displaystyle\frac{3}{\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\leq \frac{1}{3} (a+b+c)$,

allora $a b+a c+b c\geq 9 a b c.$ Also, $a+b+c=1\,\Rightarrow\,\frac{1}{3} \geq \sqrt[3]{a b c}.$

\displaystyle\begin{align} &\left(\frac{1}{a}+1\right) \left(\frac{1}{b}+1\right) \left(\frac{1}{c}+1\right)\\ &\qquad=\frac{(a+1) (b+1) (c+1)}{a b c}\\ &\qquad=\frac{a b c+a b+a c+a+b c+b+c+1}{a b c}\\ &\qquad\geq \frac{2}{a b c}+10 \geq 64. \end{align}

### Solution 4

First off, $\displaystyle\sqrt[3]{xyz}\le\displaystyle\frac{x+y+z}{3}=\frac{1}{3}.$ It follows that

\displaystyle\begin{align} \prod_{cycl}\left(1+\frac{1}{x}\right)&=1+\frac{xy+yz+zx+2}{xyz}\\ &\ge 1+\frac{3\sqrt[3]{(xyz)^2}+2}{xyz}\\ &\ge 1+3\left(\frac{1}{\sqrt[3]{xyz}}\right)+2\left(\frac{1}{\sqrt[3]{xyz}}\right)^3\\ &\ge 1+3\cdot 3+2\cdot3^2=64. \end{align}

### Solution 5

As was proved on August 31, 2008,

$a+b+b+c\ge 2\sqrt{(a+b)(b+c)}\\ b+c+c+a\ge 2\sqrt{(b+c)(c+a)}\\ c+a+a+b\ge 2\sqrt{c+a)(a+b)}.$

The product of the three is

\displaystyle\begin{align} (1+a)(1+b)(1+c)&\ge 8(a+b)(b+c)(c+a)\\ &\ge 8\cdot 2\sqrt{ab}\cdot 2\sqrt{bc}\cdot 2\sqrt{ca}\\ &=64abc. \end{align}

Hence, the required inequality. Equality is attained for $\displaystyle a=b=c=\frac{1}{3}.$

### Solution 6

\displaystyle\begin{align} LHS&=\prod_{cycl}\frac{2a+b+c}{a}\\ &=\prod_{cycl}\frac{c+a+a+b}{a}\\ &\ge\frac{8(a+b)(b+c)(c+a)}{abc}\\ &\ge\frac{8\cdot 8abc}{abc}=64. \end{align}

### Acknowledgment

The problem, with a solution, was kindly posted by Leo Giugiuc at the CutTheKnotMath facebook page. The problem and Solution 1 are by Leo Giugiuc and Marian Cucoanes. Solution 2 is by Amit Itagi; Solution 3 is by N. N. Taleb; Solutions 4,5,6 are by Kunihiko Chikaya

[an error occurred while processing this directive]