One Inequality - Two Domains

Problem 1

One Inequality - Two Domains, problem 1

Problem 2

One Inequality - Two Domains, problem 2

Solution to Problem 1

We'll start with the obvious $a^2+ab+b^2\ge\displaystyle\frac{3}{4}(a+b)^2,\,$ from which

$\displaystyle\prod_{cycl}(a^2+ab+b^2)\ge\frac{27}{64}[(a+b)(b+c)(c+a)]^2.$

On the other hand, it is easy to verify that

$\displaystyle\frac{(a+b)(b+c)(c+a)}{8}\ge\frac{(a+b+c)(ab+bc+ca)}{9}$

so that

$\displaystyle\frac{81}{64}[(a+b)(b+c)(c+a)]^2\ge (a+b+c)^2(ab+bc+ca)^2.$

Chaining the inequalities gives the desired,

$\displaystyle 3\prod_{cycl}(a^2+ab+b^2)\ge\left(\sum_{cycl}a\right)^2\cdot\left(\sum_{cycl}ab\right)^2.$

Solution to Problem 2

Note that

$\displaystyle \prod_{cycl}(a^2+ab+b^2)=\left(\prod_{cycl}a\right)^2\left(\prod_{cycl}ab\right)^2-\left(\prod_{cycl}ab\right)^3-abc\left(\prod_{cycl}a\right)^3.$

Hence we need to prove

$2(a+b+c)^2(ab+bc+ca)^2\ge 3(ab+bc+ca)^3+3abc(a+b+c)^3.$

In case $a+b+c=0,\,$ $ab+bc+ca\le 0,\,$ there's nothing to prove. So assume $a+b+c=3,\,$ $ab+bc+ca=3(1-t^2),\,$ $t\ge 0.$

Due to a result by Vo Quoc Ba Can, $\max (abc)=(1-t^2)(1+2t).\,$ Thus, suffice it to show that

$\displaystyle 2(1-t^2)^2\ge (1-t^2)^3+(1-t)^2(1+2t),$

i.e.,

$\displaystyle 2(1+t)^2\ge (1-t)(1+t)^3+1+2t.$

The latter reduces to the obvious $t^2(t^2+2t+2)\ge 0,\,$ thus completing the proof.

Illustration 1

Contour plot:

One Inequality - Two Domains, problem 1, illustration 1

Illustration 2

Region plot:

One Inequality - Two Domains, problem 1, illustration 2,2

One Inequality - Two Domains, problem 1, illustration 2,1

Acknowledgment

Both problems have been kindly posted at the CutTheKnotMath facebook page by Leo Giugiuc, along with his solutions. Problem 1 is by Gheorghe Duca; Problem 2 is by Michael Rozenberg.

The illustration 1 is by Gary Davis; Illustration 2 is by Nassim Nicholas Taleb.

 

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