An Inequality with Constraint IV

Statement

Leo Giugiuc has kindly communicated to me a problem by Voytech Jarnik with a solution by Claudia Nanuti, Diana Trailescu, Dan Sitaru and Leo Giugiuc.

Let $n\ge 2$ be an integer; real positive numbers $x_k,$ $k=1,\ldots,n,$ satisfy $\displaystyle\sum_{k=1}^{n}\frac{1}{1+x_k}=1.$ Prove that

$\displaystyle\sum_{k=1}^{n}\sqrt{x_k}\ge (n-1)\sum_{k=1}^{n}\frac{1}{\sqrt{x_k}}.$

Solution

The proof is by induction.

If $n=2$ then, easily, $x_1x_2=1$ which renders identity in the required inequality. Assume the inequality holds for $n=m\ge 2$ and consider the problem with $n=m+1,$ i.e., we assume that, for any positive $\{x_k\},$ $k=1,\ldots,m,$ with $\displaystyle\sum_{k=1}^{m}\frac{1}{1+x_k}=1$ we have $\displaystyle\sum_{k=1}^{m}\sqrt{x_k}\ge (m-1)\sum_{k=1}^{m}\frac{1}{\sqrt{x_k}}.$ We need to show that, for any positive $\{x_k\},$ $k=1,\ldots,m+1,$ with $\displaystyle\sum_{k=1}^{m+1}\frac{1}{1+x_k}=1$ we have $\displaystyle\sum_{k=1}^{m+1}\sqrt{x_k}\ge m\sum_{k=1}^{m+1}\frac{1}{\sqrt{x_k}}.$

Denote $\displaystyle\frac{1}{1+x_k}=a_k.$ We have $a_k\in (0,1)$ and $\sum_{k=1}^{m+1}a_k=1.$ In terms of $\{a_k\}$ we have to prove that $\displaystyle\sum_{k=1}^{m+1}\sqrt{\frac{1-a_k}{a_k}}\ge m\sum_{k=1}^{m+1}\sqrt{\frac{a_k}{1-a_k}}.$

By the induction hypothesis, $\displaystyle\sum_{k=1,k\ne i}^{m+1}\sqrt{\frac{1-a_k-a_i}{a_k}}\ge m\sum_{k=1,k\ne i}^{m+1}\sqrt{\frac{a_k}{1-a_k-a_i}},$ $1\le i\le m+1.$ Summing up and rearranging we get

$\displaystyle\sum_{k=1}^{m+1}\sum_{i=1,i\ne k}^{m+1}\sqrt{\frac{1-a_k-a_i}{a_k}}\ge m\sum_{k=1}^{m+1}\sum_{i=1,i\ne k}^{m+1}\sqrt{\frac{a_k}{1-a_k-a_i}}.$

Now, apply Jensen's inequality with the convex function $f(x)=\sqrt{x}:$

$\displaystyle\sqrt{m(m-1)\frac{1-a_k}{a_k}}\ge\sum_{i=1,i\ne k}^{m+1}\sqrt{\frac{1-a_k-a_i}{a_k}}$

and to the concave function $\displaystyle f(x)=\frac{1}{\sqrt{x}}:$

$\displaystyle\sum_{i=1,i\ne k}^{m+1}\sqrt{\frac{1-a_k-a_i}{a_k}}\ge m\sqrt{\frac{m}{m-1}}\sqrt{\frac{a_k}{1-a_k}}.$

The combination of the two inequalities gives

$\displaystyle\sqrt{m(m-1)}\sum_{k=1}^{m+1}\sqrt{\frac{1-a_k}{a_k}}\ge (m-1)m\sqrt{\frac{m}{m-1}}\sum_{k=1}^{m+1}\sqrt{\frac{a_k}{1-a_k}}$

which simplifies to

$\displaystyle\sum_{k=1}^{m+1}\frac{1-a_k}{a_k}\ge m\sum_{k=1}^{m+1}\sqrt{\frac{a_k}{1-a_k}}.$

This is the required inequality.

 

|Contact| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62044945

Search by google: