An Inequality with Constraint VI


Let $0\lt x_k,$ $k=1,\dots,n$ such that $\displaystyle\sum_{k=1}^{n}x_k=1,$ $n\ge 1.$ Then

$\displaystyle\prod_{k=1}^{n}\frac{1+x_k}{x_k}\ge \prod_{k=1}^{n}\frac{n-x_k}{1-x_k}.$


Consider function $\displaystyle f(x)=\ln\left(1+\frac{1}{x}\right).$ The function is convex on $(0,1).$ By Jensen's inequality, for any fixed $k=1,2,\ldots,n,$

$\displaystyle\frac{1}{n-1}\sum_{i=1,i\ne k}^{n}\ln\left(1+\frac{1}{x_i}\right)\ge \ln\left(1+\frac{n-1}{\displaystyle\sum_{i=1,i\ne k}^{n}x_i}\right)=\ln\left(\frac{n-x_{k}}{1-x_k}\right).$

In other words, $\displaystyle\prod_{i=1,i\ne k}^{n}\left(1+\frac{1}{x_i}\right)\ge\left(\frac{n-x_k}{1-x_k}\right)^{n-1}.$ The desired inequality is obtained on multiplying all $n\;$ of these inequalities up.


  1. Xu Jiagu, Lecture Notes on Mathematical Olympiad Courses, v 8, (For senior section, v 2), World Scientific, 2012, 88-89


Inequalities with the Sum of Variables as a Constraint

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