# An Inequality in Two Or More Variables

### 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 $n.$