Ishaan Booker

Answered

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:

$(\frac{1}{{t}_{1}}-1)(\frac{1}{{t}_{2}}-1)...(\frac{1}{{t}_{n+1}}-1)\ge {n}^{n+1}$

Which is equavalent to the following:

$(\frac{{t}_{2}+{t}_{3}+...+{t}_{n+1}}{{t}_{1}})(\frac{{t}_{1}+{t}_{3}+...+{t}_{n+1}}{{t}_{2}})...(\frac{{t}_{1}+{t}_{2}+...+{t}_{n}}{{t}_{n+1}})\ge {n}^{n+1}$

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.

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:

$(\frac{1}{{t}_{1}}-1)(\frac{1}{{t}_{2}}-1)...(\frac{1}{{t}_{n+1}}-1)\ge {n}^{n+1}$

Which is equavalent to the following:

$(\frac{{t}_{2}+{t}_{3}+...+{t}_{n+1}}{{t}_{1}})(\frac{{t}_{1}+{t}_{3}+...+{t}_{n+1}}{{t}_{2}})...(\frac{{t}_{1}+{t}_{2}+...+{t}_{n}}{{t}_{n+1}})\ge {n}^{n+1}$

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.

Answer & Explanation

frisiao

Expert

2022-07-21Added 13 answers

By AM-GM

$\prod _{i=1}^{n+1}\frac{{x}_{i}}{{x}_{i}+1}=\prod _{i=1}^{n+1}\sum _{k\ne i}\frac{1}{{x}_{k}+1}\ge \prod _{i=1}^{n+1}\frac{n}{\sqrt[n]{\prod _{k\ne i}(1+{x}_{k})}}=\frac{{n}^{n+1}}{\prod _{i=1}^{n+1}(1+{x}_{i})}$

and we are done!

$\prod _{i=1}^{n+1}\frac{{x}_{i}}{{x}_{i}+1}=\prod _{i=1}^{n+1}\sum _{k\ne i}\frac{1}{{x}_{k}+1}\ge \prod _{i=1}^{n+1}\frac{n}{\sqrt[n]{\prod _{k\ne i}(1+{x}_{k})}}=\frac{{n}^{n+1}}{\prod _{i=1}^{n+1}(1+{x}_{i})}$

and we are done!

enmobladatn

Expert

2022-07-22Added 6 answers

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.

$\frac{{t}_{2}+{t}_{3}+...+{t}_{n+1}}{n}\ge \sqrt[n]{{t}_{2}{t}_{3}...{t}_{n+1}}=\frac{A}{\sqrt[n]{{t}_{1}}}$

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.

Most Popular Questions