An Inequality Involving Arithmetic And Geometric Means

Problem

An Inequality Involving Arithmetic And Geometric Means

Solution 1

By the AM-GM inequality, applied repeatedly,

$\begin{align} &a^4+b^4\ge 2a^2b^2\\ &b^4+c^4\ge 2b^2c^2\\ &c^4+a^4\ge 2c^2a^2 \end{align}\Bigg\}\Rightarrow a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2. $

$\begin{align} &a^2b^2+b^2c^2\ge 2ab^2c\\ &b^2c^2+c^2a^2\ge 2abc^2\\ &c^2a^2+a^2b^2\ge 2a^2bc \end{align}\Bigg\}\Rightarrow a^2b^2+b^2c^2+c^2a^2\ge abc(a+b+c). $

It follows that $a^4+b^4+c^4\ge abc(a+b+c);$ and, subsequently,

$\displaystyle\begin{align} &a^4+b^4+c^4+abcd\ge abc(a+b+c+d)&\Rightarrow\\ &\sum_{cycl}\frac{1}{a^4+b^4+c^4+abcd}\le\frac{1}{a+b+c+d}\sum_{cycl}\frac{1}{abc}\\ &\qquad\qquad\qquad=\frac{1}{a+b+c+d}\sum_{cycl}\frac{a}{abcd}=\frac{1}{abcd}. \end{align}$

Equality is attained for $a=b=c=d.$

Solution 2

We'll prove a more general statement:

Let $a_1,a_2,\ldots,a_n$ be positive real numbers. Set, for an integer $m,$ $\displaystyle S^{(m)}=\sum_{k=1}^na_k^m$ and $\displaystyle P=\prod_{k=1}^na_k.$ Denote also $S_k^{(m)}=S^{(m)}-a_k^m,$ $k=1,2,\ldots,n.$ Then

$\displaystyle\sum_{k=1}^n\frac{1}{S_k^{(n)}+P}\le\frac{1}{P}.$

Indeed, by Jensen's inequality

$\displaystyle\begin{align}\frac{S_k^{(n)}}{n-1}&\ge\left(\frac{S_k^{(1)}}{n-1}\right)^n\\ &=\frac{1}{n-1}S_k^{(1)}\left(\frac{S_k^{(1)}}{n-1}\right)^{n-1}\\ &=\frac{1}{n-1}S_k^{(1)}\left(\sqrt[n-1]{\frac{P}{a_k}}\right)^{n-1}, \end{align}$

implying $\displaystyle S_k^{(n)}\ge S_k^{(1)}\frac{P}{a_k}.$ Thus, summing up,

$\displaystyle\begin{align} \sum_{k=1}^n\frac{1}{S_k^{(n)}+P}&\le \sum_{k=1}^n\frac{1}{\displaystyle P\left(1+\frac{S_k^{(1)}}{a_k}\right)}\\ &=\frac{1}{P}\sum_{k=1}^n\frac{a_k}{a_k+S_k^{(1)}}\\ &=\frac{1}{P}\sum_{k=1}^n\frac{a_k}{S^{(1)}}=\frac{1}{P}. \end{align}$

Solution 3

$\displaystyle \begin{align} &\frac{a^4+b^4+c^4}{3} \ge \left(\frac{a+b+c}{3}\right)^4\\ &a^4+b^4+c^4 \ge \frac{(a+b+c)^4}{27}\\ &\frac{a+b+c}{3} \ge \sqrt[3]{abc}\\ &a+b+c\ge 3\sqrt[3]{abc}\\ &a^4+b^4+c^4 \ge \frac{(a+b+c)(a+b+c)^3}{27}\\ &a^4+b^4+c^4 \ge (a+b+c)\frac{(3\sqrt[3]{abc})^3}{27}\\ &a^4+b^4+c^4 \ge (a+b+c)abc\\ &a^4+b^4+c^4+abcd \ge abc(a+b+c)+abcd\\ &d(a^4+b^4+c^4+abcd) \ge abcd(a+b+c+d)\\ &\frac{abcd}{a^4+b^4+c^4+abcd} \ge \frac{d}{a+b+c+d}\\ &\sum_{cycl}\frac{abcd}{a^4+b^4+c^4+abcd} \ge \frac{a+b+c+d}{a+b+c+d}=1 \end{align}$

Acknowledgment

The problem has been kindly posted at the CutTheKnotMath facebook page by Diego Alvariz, yet in 2015.

Solution 1 is by Trinh Hào Quang; Solution 2 is by Regragui El Khammal; Solution 3 is by Sourabh Dubey.

 

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