A Cyclic Inequality in Three Variables XVI

Problem

A  Cyclic Inequality  in  Three Variables  XVI

Solution 1

$\displaystyle\begin{align} &[(1+a)+b(1-a)]^2\ge 0\;\Longleftrightarrow\\ &(1+a)^2+2b(1-a^2)+b^2(1-a)^2\ge 0\;\Longleftrightarrow\\ &1+2a+a^2+2b-2ba^2+b^2-2b^2a+b^2a^2\geq 0\;\Longleftrightarrow\\ &1+a^2+b^2+a^2b^2+2(a+b-a^2b-ab^2)\geq 0\;\Longleftrightarrow\\ &(1+a^2)(1+b^2)\geq 2\Bigr(ab(a+b)-(a+b)\Bigr)\;\Longleftrightarrow\\ &2(a+b)(ab-1)\leq (1+a^2)(1+b^2)\;\Longleftrightarrow\\ \end{align}$

(1)

$2(a+b)(1-ab)\geq -(1+a^2)(1+b^2).$

Further,

$\displaystyle\begin{align} &[(1-a)-b(1+a)]^2\geq 0\;\Longleftrightarrow\\ &(1-a)^2-2b(1-a^2)+b^2(1+a)^2\geq 0\;\Longleftrightarrow\\ &1-2a+a^2-2b+2ba^2+b^2+2ab^2+a^2b^2\geq 0\;\Longleftrightarrow\\ &1+a^2+b^2+a^2b^2-2(a+b-ab^2-a^2b)\geq 0\;\Longleftrightarrow\\ &(1+a^2)(1+b^2)\geq 2\Bigr(a+b-ab(a+b)\Bigr)\;\Longleftrightarrow\\ \end{align}$

(2)

$2(a+b)(1-ab)\leq (1+a^2)(1+b^2).$

From (1),(2) it follows that

(3)

$2|(a+b)(1-ab)|\leq (1+a^2)(1+b^2).$

Simialrly,

(4)

$2|(b+c)(1-bc)|\leq (1+b^2)(1+c^2).$


(5)

$2|(c+a)(1-ca)|\leq (1+c^2)(1+a^2).$

In (3) equality is attained for $a=0;\,b=1\,$ or $a=1;\,b=0;\,$ similarly, for (4) and (5). Thus, the sum_{cycl} of the three inequality is strict:

$\displaystyle\begin{align} &2\sum_{cycl} |(a+b)(1-ab)|\lt\sum_{cycl} (1+a^2+b^2+a^2b^2)\\ &=3+2(a^2+b^2+c^2)+\sum_{cycl} a^2b^2\\ &\lt 3+2(a^2+b^2+c^2)+\sum_{cycl} a^4 \end{align}$

Dividing by $2,$

$\displaystyle\sum_{cycl} |(a+b)(1-ab)|\lt \frac{3}{2}+a^2+b^2+c^2+\frac{1}{2}\sum_{cycl} a^4.$

Solution 2

We write $|(a+b)(1-ab)|\,$ as $\sqrt{(a+b)^2(1-ab)^2}.\,$ By the AM-GM inequality,

$\displaystyle\begin{align} \sqrt{(a+b)^2(1-ab)^2}&\le\frac{1}{2}((a-b)^2+(1-ab)^2)\\ &=\frac{1}{2}(1+a^2+b^2+a^2b^2). \end{align}$

Alors,

$\displaystyle\begin{align} \sum_{cycl}\sqrt{(a+b)^2(1-ab)^2}&\le\frac{1}{2}\sum_{cycl}(1+a^2+b^2+a^2b^2)\\ &=\frac{3}{2}+\sum_{cycl}a^2+\frac{1}{2}\sum_{cycl}a^2b^2. \end{align}$

However, $\displaystyle\sum_{cycl}a^2b^2\le\sum_{cycl}a^4,\,$ by the Rearrangement inequality.

Acknowledgment

This problem with the solution (Solution 1) has been kindly communicated to me by Dan Sitaru, all on a tex file. Solution 2 is by N. N. Taleb.

 

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