Ishaan Booker

2022-07-20

An Elementary Inequality Problem
Given $n$ positive real numbers ${x}_{1},{x}_{2},...,{x}_{n+1}$, suppose:
$\frac{1}{1+{x}_{1}}+\frac{1}{1+{x}_{2}}+...+\frac{1}{1+{x}_{n+1}}=1$
Prove:
${x}_{1}{x}_{2}...{x}_{n+1}\ge {n}^{n+1}$
I have tried the substitution
${t}_{i}=\frac{1}{1+{x}_{i}}$
The problem thus becomes:
Given
${t}_{1}+{t}_{2}+...+{t}_{n+1}=1,{t}_{i}>0$
Prove:
$\left(\frac{1}{{t}_{1}}-1\right)\left(\frac{1}{{t}_{2}}-1\right)...\left(\frac{1}{{t}_{n+1}}-1\right)\ge {n}^{n+1}$
Which is equavalent to the following:
$\left(\frac{{t}_{2}+{t}_{3}+...+{t}_{n+1}}{{t}_{1}}\right)\left(\frac{{t}_{1}+{t}_{3}+...+{t}_{n+1}}{{t}_{2}}\right)...\left(\frac{{t}_{1}+{t}_{2}+...+{t}_{n}}{{t}_{n+1}}\right)\ge {n}^{n+1}$
From now on, I think basic equality (which says arithmetic mean is greater than geometric mean) can be applied but I cannot quite get there. Any hints? Thanks in advance.

frisiao

Expert

By AM-GM
$\prod _{i=1}^{n+1}\frac{{x}_{i}}{{x}_{i}+1}=\prod _{i=1}^{n+1}\sum _{k\ne i}\frac{1}{{x}_{k}+1}\ge \prod _{i=1}^{n+1}\frac{n}{\sqrt[n]{\prod _{k\ne i}\left(1+{x}_{k}\right)}}=\frac{{n}^{n+1}}{\prod _{i=1}^{n+1}\left(1+{x}_{i}\right)}$
and we are done!

Expert

From last, let $\sqrt[n]{{t}_{1}{t}_{2}...{t}_{n+1}}=A$ with $AM-GM$
$\frac{{t}_{2}+{t}_{3}+...+{t}_{n+1}}{n}\ge \sqrt[n]{{t}_{2}{t}_{3}...{t}_{n+1}}=\frac{A}{\sqrt[n]{{t}_{1}}}$
so for the first
${t}_{2}+{t}_{3}+...+{t}_{n+1}\ge \frac{nA}{\sqrt[n]{{t}_{1}}}$
and for the rest take like this.

Do you have a similar question?