A Refinement of Turkevich's Inequality


A Refinement of Turkevich's Inequality

Solution 1

Turkevich's inequality states:

If $a,b,c,d\ge 0,\,$ then

$a^2+b^2+c^2+d^2+2\sqrt{abcd}\ge ab+bc+cd+da+ac+bd.$

Why is the problem a refinement? Because, by the AM-GM inequality, $\displaystyle 2\sqrt{abcd}\ge\frac{32abcd}{(a+b+c+d)^2}.$

Assume, WLOG, $a+b+c+d=4.\,$ Clearly, $\displaystyle \sum_{all}ab=6(1-t^2),\,$ with $t\in [0,1].\,$ From here, $\displaystyle \sum_{cycl}a^2=4(1+3t^2).\,$ We need to prove that $2(1+3t^2)+abcd\ge 3(1-t^2).$

I (AB: Leo Giugiuc) have shown very often that

$\min (abcd)=\begin{cases} (1+t)^3(1-3t),& if\;\displaystyle 0\le t\le \frac{1}{3}\\ 0,& if\; \displaystyle \frac{1}{3}\le t \le 1. \end{cases}$

$\mathbf{Case\,1:\,\displaystyle 0\le t\le \frac{1}{3}}$

Suffice it to show that $2(1+3t^2)+(1+t^3)(1-3t)\ge 3(1-t^2),\,$ which is equivalent to $(1=t)^3(1-3t)\ge(1+3t)(1-3t),\,$ or, $(1+t)^3\ge 1+3t.\,$ The latter is obviously true.

$\mathbf{Case\,2:\,\displaystyle \frac{1}{3}\le t \le 1}$

Suffice it to show that $2(1+t^2)\ge 3(1-t^2),\,$ which is equivalent to $0\ge (1+3t)(1-3t),\,$ which is true.

Let's remark that equality holds at $(a,a,a,a)\,$ and $(a,a,a,0)\,$ and permutations, $a\gt 0.$

Solution 2

We start with an observation that

$\displaystyle\begin{align} &\left(\sum_{sym}a^2\right)\left(\sum_{sym}a\right)^2+32abcd-\left(\sum_{sym}a\right)\left(\sum_{sym}ab\right)\\ &\qquad\qquad+\sum_{sym}ab(a^2+b^2)-[(ab+cd)(ac+bd)\\ &\qquad\qquad\qquad+(ab+cd)(ad+bc)+(ad+bc)(ac+bd)]. \end{align}$

We thus have that the required inequality is equivalent to


$\displaystyle\begin{align} &\sum_{sym}a^4+20abcd+\sum_{sym}ab(a^2+b^2)\ge 3[(ab+cd)(ac+bd)\\ &\qquad\qquad\qquad+(ab+cd)(ad+bc)+(ad+bc)(ac+bd)]. \end{align}$

We are going to make use of another of Turkevich's inequalities:


$\displaystyle \sum_{sym}a^4+2abcd\ge\sum_{sym}a^2b^2.$

Using the AM-GM inequality,


$\displaystyle \sum_{sym}ab(a^2+b^2)\ge 2\sum_{sym}a^2b^2.$

From (2)&(3),

$\displaystyle \begin{align} &\sum_{sym}a^4+20abcd+\sum_{sym}ab(a^2+b^2)\ge 18abcd+\sum_{sym}a^2b^2+2\sum_{sym}a^2b^2\\ &\qquad\qquad=3[(ab+cd)^2+(ad+bc)^2+(ac+bd)^2]\\ &\qquad\qquad\ge 3[(ab+cd)(ac+bd)+(ab+cd)(ad+bc)+(ad+bc)(ac+bd)] \end{align}$

so that (1) does imply the required inequality.


The problem and Solution 1 are by Leo Giugiuc who kindly posted the problem at the CutTheKnotMath facebook page and then also mailed me his solution. I could not be more appreciative. Solution 2 is by Marian (Gabi Cuc) Cucoaneş.


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