# An Inequality in Two Or More Variables

### Problem

### Solution 1

$\displaystyle \begin{align} &\frac{a}{1+a}+\frac{b}{(1+a)(1+b)}+\frac{c}{(1+a)(1+b)(1+c)}\\ &\qquad=\frac{a(1+b)(1+c)+b(1+c)+c}{(1+a)(1+b)(1+c)}\\ &\qquad=\frac{(1+c)(a+ab+b+1)-1}{(1+a)(1+b)(1+c)}\\ &\qquad=\frac{(1+c)(1+b)(1+c)-1}{(1+a)(1+b)(1+c)}\\ &\qquad=1-\frac{1}{(1+a)(1+b)(1+c)}\overbrace{\geq}^{AM-GM}1-\frac{1}{2\sqrt{a}\cdot 2\sqrt{b}\cdot 2\sqrt{c}}\\ &\qquad=1-\frac{1}{8\sqrt{abc}}=1-\frac{1}{8}=\frac{7}{8} \end{align}$

### A little extra

The above statement and proof extend easily to a number $n\ge 2\,$ of variables:

Prove that if $a_k\gt 0,\,$ $k=1,2,\ldots,n\,$ and $\displaystyle \prod_{k=1}^na_k=1$ then

$\displaystyle \sum_{k=1}^n\left(a_k\prod_{i=1}^k\frac{1}{1+a_i}\right)\ge\frac{2^n-1}{2^n}.$

The key is the identity derived in the above proof:

$\displaystyle \sum_{k=1}^n\left(a_k\prod_{i=1}^k\frac{1}{1+a_i}\right)=1-\prod_{i=1}^n\frac{1}{1+a_i}.$

The identity can be established by induction. Let $\displaystyle S_n=\sum_{k=1}^n\left(a_k\prod_{i=1}^k\frac{1}{1+a_i}\right)\,$ and $\displaystyle P_n=1-\prod_{i=1}^n\frac{1}{1+a_i}.\,$ Then

$\displaystyle\begin{align}S_{n+1}-S_{n}&=a_{n+1}\prod_{i=1}^{n+1}\frac{1}{1+a_i},\,\text{whereas}\\ P_{n+1}-P_{n}&=\prod_{i=1}^{n}\frac{1}{1+a_i}-\prod_{i=1}^{n+1}\frac{1}{1+a_i}\\ &=\prod_{i=1}^{n}\frac{1}{1+a_i}\left(1-\frac{1}{1+a_{n+1}}\right)\\ &=\prod_{i=1}^{n}\frac{1}{1+a_i}\left(\frac{a_{n+1}}{1+a_{n+1}}\right)\\ &=a_{n+1}\prod_{i=1}^{n+1}\frac{1}{1+a_i}. \end{align}$

The identity holds for $n=3\,$ (and obviously for $n=1\,$ and $n=2,)$ hence, it holds for any larger $n.$

### Solution 2

Multiplying out: $abc+(ab+c)+(bc+a)+(ca+b) \ge 7,\,$ i.e., $\displaystyle 1+(1/c+c)+(1/a+a)+(1/b+b) \ge 7\, which follows from the AM-GM inequality applied to each pair of parentheses.

### Solution 3

$\displaystyle f=\frac{c}{(a+1) (b+1) (c+1)}+\frac{b}{(a+1) (b+1)}+\frac{a}{a+1}$

By rearranging the terms,

$\displaystyle\begin{align} f&=1-\frac{1}{(a+1) (b+1) (c+1)}\\ &=1-\frac{1}{1+a+b+c+ab+bc+ca+abc}\\ &\geq 1-\frac{27}{(a+b+c+3)^3}. \end{align}$

By the AM-GM inequality,

$\displaystyle 1+a+b+c+ab+bc+ca+abc\ge 8*sqrt[8]{a^4b^4c^4}=8.$

Can be generalized to $n$ summands.

### Acknowledgment

Dan Sitaru has kindly posted the problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. He later mailed his solution on a LaTex file - something I appreciate greatly. Solution 2 is by Amit Itagi; Solution 3 is by N. N. Taleb.

Roland van Gaalen has observed that for a sequence $a_1, a_2, \ldots\,$ that satisfy $\displaystyle \prod_{i}a_i=1,\,$ the sum $\displaystyle \sum_{k=1}^{\infty}\left(a_k\prod_{i=1}^k\frac{1}{1+a_i}\right)=1.$

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