Second Pair of Twin Inequalities: Twin 2

Problem

Solution 1

Let $S$ denote the set of all subsets of distinct indices from $1$ to $n.$ $(|S|=2^n).$ If $\mathbf{j}=(j_1,j_2,\ldots,j_k)\in S,$ we'll write $a_{\mathbf{j}}=a_{j_1}a_{j_2}\cdot\ldots\cdot a_{j_k}.$ $S=\bigcup S_k,$ where $S_k$ consists of the sequences of length $k.$ Note that $\displaystyle |S_k|={n\choose k}.$ Further

$\displaystyle\prod_{i=1}^n\left(\frac{1}{a_i}+1\right)=\sum_{k=0}^n\sum_{\mathbf{j}\in S_k}\frac{1}{a_{\mathbf{j}}}.$

Every index $a_j$ appears in $\displaystyle {n-1\choose k-1}$ sequences from $S_k.$ So that, applying the AM-GM inequality,

$\displaystyle \begin{align} \sum_{\mathbf{j}\in S_k}\frac{1}{a_{\mathbf{j}}}&\ge{n\choose k}\sqrt[{n\choose k}]{P^{{n-1\choose k-1}}}\\ &={n\choose k}P^{\frac{k}{n}}, \end{align}$

where $\displaystyle P=\frac{1}{a_1a_2\cdot\ldots\cdot a_n}.$ By the AM-GM inequality,

$\displaystyle P=\frac{1}{a_1a_2\cdot\ldots\cdot a_n}\ge\left(\frac{n}{a_1+a_2+\ldots+a_n}\right)^n=n^n.$

It follows that

$\displaystyle\begin{align} \prod_{i=1}^n\left(\frac{1}{a_i}+1\right)&\ge\sum_{k=0}^n{n\choose k}P^{\frac{k}{n}}=\left(1+P^{\frac{1}{n}}\right)^n\\ &\ge (n+1)^n. \end{align}$

Solution 2

$\displaystyle\begin{align} \prod_{i=1}^n\left(\frac{1}{a_i}+1\right)&=\prod_{i=1}^n\left(\frac{\displaystyle \sum_{j=1}^na_j+a_i}{a_i}\right)\\ &=\frac{\displaystyle \prod_{i=1}^n\left(\sum_{j=1}^na_j+a_i\right)}{\displaystyle \prod_{i=1}^na_i}. \end{align}$

Now, by the AM-GM inequality and by the symmetry of inclusion of the terms $a_i$,

$\displaystyle \prod_{i=1}^n\left(\sum_{j=1}^na_j+a_i\right)\ge (n+1)^n\left(\prod_{i=1}^na_j\right)^{\frac{n+1}{n+1}}$

from which the required inequality follows.

Solution 3

Function $\displaystyle f(x)=\ln\left(1+\frac{1}{x}\right)$ is convex, so that, by Jensen's inequality,

$\displaystyle \begin{align} \frac{1}{n}\sum_{i=1}^n\ln\left(1+\frac{1}{a_i}\right)&\ge\ln\left(1+\frac{n}{\sum_{i=1}^na_i}\right)\\ &=\ln(1+n), \end{align}$

implying $\displaystyle \prod_{i=1}^n\left(1+\frac{1}{a_i}\right)\ge (n+1)^n.$

Solution 4

Assume we have $0\lt x\le y\le 1-x.$ Then

$\displaystyle \left(\frac{1}{x}+1\right)\left(\frac{1}{y}+1\right) \gt \left(\frac{2}{x+y}+1\right)^2.$

Indeed, this is equivalent to $\displaystyle ((y - x)^2 (x + y + 1))/(x y (x + y)^2)\ge 0.$ Below is an illustration with wolframalpha:

(1/x+1)(1/y+1) not less than (2/(x+y)+1)^2

This justifies a step at a time process whereby we modify the terms of the sequence ${\a_i\}$ two at a time until (or in the limit) all become equal.

Acknowledgment

This is an adaptation of a problem from R. Honsberger's In Polya's Footsteps (MAA, 1997, pp 26-28.) Honsberger credits Solution 2 to Paul Schellenberg. Solution 3 is by N. N. Taleb; Solution 4 is by Christopher D. Long.

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