Monthly Problem 11199

Introduction

It's so happened that I've posting lately various inequalities, supplied mostly by a Romanian team of Leonard Giugiuc, Daniel Sitaru, Dorin Marghidanu, and Marian Dincu. This activity on twitter.com has attracted attention of Marc Chamberland (TPointMath, https://www.youtube.com/user/TippingPointMath.) Marc has suggested that a paper by an Austrian colleague of his, Manuel Kauers, that deals with automatic solution of inequalities, might be of interest.

One sample inequality discussed in the paper was the Monthly problem 11199. When submitted to CAD (Cylindrical Algebraic Decomposition), it engendered the output "true." While this is a definite achievement, the algorithm underlying CAD is doubly exponential and at present cannot compete with human ingenuity.

Thus I consider below the problem 11199 (Yakub Aliyev, Baku State University, Baku, Azerbaijan) and the solution (Monthly, August-September 2007, p 649) by David B. Leep (University of Kentucky, Lexington, KY.) The solution leads naturally to a modified problem whose solution has been devised by Leo Giugiuc.

Problem 11199

Monthly Problem 11199

Solution to Problem 11199

Introduce function $s_1=a+b+c,\;$ $s_2=ab+bc+ca,\;$ and $s_3=abc.\;$ Observe that

$\begin{align} s_1^3s_2 &+ 48s_2s_3 - 25s_1^2s_3\\ &= a(b - c)^2(3a - b - c)^2 + b(c - a)^2(3b - c - a)^2 + c(a - b)^2(3c - a - b)^2\\ &\ge 0. \end{align}$

Since $s_1=1,\;$ this gives $s_2 + 48s_2s_3 - 25s_1^2s_3\ge 0\;$ which rearranges into the required inequality.

Modification

Note that the same expression that was the key to solving the problem 11199 applies with the same ease to solve an apparently different problem:

For positive $a,b,c,\;$ with $abc=1,\;$ prove that

$\displaystyle ab+bc+ca\ge\frac{25(a+b+c)^2}{(a+b+c)^3+48}.$

Homogenization, or Solution 2 to the Modification

Since $abc=1,\;$ we may as well prove $\displaystyle\frac{ab+bc+ca}{abc}\ge\frac{25(a+b+c)^2}{(a+b+c)^3+48abc},\;$ $a,b,c\gt 0,\;$ without additional constraints. One way of doing that is by introducing, perhaps paradoxically, a constraint. Let, WLOG, $a+b+c=3.$

Then, by a theorem disovered independetly by Leo Giugiuc, Vo Quoc Ba Can, and probably others, there is $t\in [0,1)\;$ such that $ab+bc+ca\le 3(1-t^2)\;$ and $abc\le(1-t)^2(1+2t).\;$ Our inequality is equivalent to $9(1-t^2)-(9+16t^2)abc\ge 0.\;$ Thus, suffice it to prove a stronger inequality:

$9(1-t^2)-(9+16t^2)(1-t)^2(1+2t)\ge 0.$

But the latter can be rearranged into $2(1-t)t^2(4t-1)^2\ge 0\;$ which is clearly true for $t\in [0,1).$

Observe that the equality holds if $a=b=c=1\;$ or $\displaystyle a=b=\frac{3}{4}\;$ and $\displaystyle c=\frac{3}{2}.\;$ More generally, it holds for $a=b=c\gt 0\;$ or $c=2a=2b,\;$ $a,b,c\gt 0,\;$ and the permutations.

Homogenization, or Solution 2 to the Problem 11199

Note that to prove $\displaystyle\frac{ab+bc+ca}{abc}\ge\frac{25(a+b+c)^2}{(a+b+c)^3+48abc},\;$ we could have taken $a+b+c=1\;$ which would have produced the original problem 11199. We would then have $\displaystyle ab+bc+ca\le \frac{1-t^2}{3}\;$ and $\displaystyle abc\le\frac{(1-t)^2(1+2t)}{27},\;$ but, otherwise the reasoning (i.e., algebra) would be about the same.

Thus we can see that

Leo Giugiuc's solution of the modification also solves the original problem.
Moreover, via the process of the homogenization, the two problems are not just equivalent, they are two views of one and the same problem.

Inequalities with the Sum of Variables as a Constraint

An Inequality for Grade 8 $\left(\displaystyle\frac{1-x_1}{1+x_1}\cdot\frac{1-x_2}{1+x_2}\cdot\ldots\cdot\frac{1-x_n}{1+x_n}\ge\frac{1}{3}\right)$
An Inequality with Constraint $((x+1)(y+1)(z+1)\ge 4xyz)$
An Inequality with Constraints II $\left(\displaystyle abc+\frac{2}{ab+bc+ca}=p+\frac{2}{q}\ge q-2+\frac{2}{q}\right)$
An Inequality with Constraint V $\left(\displaystyle\prod_{k=1}^{n}x_k^{1/x_k}\le \frac{1}{n^{n^2}}\right)$
An Inequality with Constraint VI $\left(\displaystyle\prod_{k=1}^{n}\frac{1+x_k}{x_k}\ge \prod_{k=1}^{n}\frac{n-x_k}{1-x_k}\right)$
An Inequality with Constraint XI $(\sqrt{5a+4}+\sqrt{5b+4}+\sqrt{5c+4} \ge 7)$
Monthly Problem 11199 $\left(\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{25}{1+48abc}\right)$
Problem 11804 from the AMM $(10|x^3 + y^3 + z^3 - 1| \le 9|x^5 + y^5 + z^5 - 1|)$
Sladjan Stankovik's Inequality With Constraint $\left(abc+bcd+cda+dab-abcd\le\displaystyle \frac{27}{16}\right)$
Sladjan Stankovik's Inequality With Constraint II $(a^4+b^4+c^4+d^2+4abcd\ge 8)$
An Inequality with Constraint V $\left(\displaystyle\prod_{k=1}^{n}x_k^{1/x_k}\le \frac{1}{n^{n^2}}\right)$
An Inequality with Constraint VI $\left(\displaystyle\prod_{k=1}^{n}\frac{1+x_k}{x_k}\ge \prod_{k=1}^{n}\frac{n-x_k}{1-x_k}\right)$
An Inequality with Constraint XII $(abcd\ge ab+bc+cd+da+ac+bd-5)$
An Inequality with Constraint XIII $((3a-bc)(3b-ca)(3c-ab)\le 8a^2b^2c^2)$
Inequalities with Constraint XV and XVI $\left(\displaystyle\frac{a^2}{\sqrt{b^2+4}}+\frac{b^2}{\sqrt{c^2+4}}+\frac{c^2}{\sqrt{a^2+4}}\gt\frac{3}{5}\right)$ and $\left(\displaystyle\frac{a^2}{\sqrt{b^4+4}}+\frac{b^2}{\sqrt{c^4+4}}+\frac{c^2}{\sqrt{a^4+4}}\gt\frac{3}{5}\right)$
An Inequality with Constraint XVII $(a^3+b^3+c^3\ge 0)$
An Inequality with Constraint in Four Variables $\left(\displaystyle\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b}\ge\frac{1}{8}\right)$
An Inequality with Constraint in Four Variables II $(a^3+b^3+c^3+d^3 + 6abcd \ge 10)$
An Inequality with Constraint in Four Variables III $\left(\displaystyle\small{abcd+\frac{15}{2(ab+ac+ad+bc+bd+cd)}\ge\frac{9}{a^2+b^2+c^2+d^2}}\right)$
An Inequality with Constraint in Four Variables IV $\left(\displaystyle 27+3(abc+bcd+cda+dab)\ge\sum_{cycl}a^3+54\sqrt{abcd}\right)$
Inequality with Constraint from Dan Sitaru's Math Phenomenon $\left(\displaystyle b+2a+20\ge 2\sum_{cycl}\frac{a^2+ab+b^2}{a+b}\ge b+2c+20\right)$
An Inequality with a Parameter and a Constraint $\left(\displaystyle a^4+b^4+c^4+\lambda abc\le\frac{\lambda +1}{27}\right)$
Cyclic Inequality with Square Roots And Absolute Values $\left(\displaystyle \prod_{cycl}\left(\sqrt{a-a^2}+\frac{1}{2\sqrt{2}}|3a-1|\right)\ge\frac{1}{6\sqrt{6}}\prod_{cycl}\left(\sqrt{a}+\frac{1}{\sqrt{3}}\right)\right)$
From Six Variables to Four - It's All the Same $\left(\displaystyle \frac{5}{2}\le a^2+b^2+c^2+d^2\le 5\right)$
Michael Rozenberg's Inequality in Three Variables with Constraints $\left(\displaystyle 4\sum_{cycl}ab(a^2+b^2)\ge\sum_{cycl}a^4+5\sum_{cycl}a^2b^2+2abc\sum_{cycl}a\right)$
Michael Rozenberg's Inequality in Two Variables $\left(\displaystyle \sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6\right)$
Dan Sitaru's Cyclic Inequality in Three Variables II $\left(\displaystyle \sum_{cycl}\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}\geq \frac{9}{12-2(ab+bc+ca)}+3\right)$
Dan Sitaru's Cyclic Inequality in Three Variables IV $\left(\displaystyle \sum_{cycl}\frac{(x+y)z}{\sqrt{4x^2+xy+4y^2}}\le 2\right)$
Dan Sitaru's Cyclic Inequality in Three Variables VI $\left(\displaystyle \sum_{cycl}\left[\sqrt{a(a+2b)}+\sqrt{b(b+2a)}\,\right]\le 6\sqrt{3}\right)$
An Inequality with Arbitrary Roots $\left(\displaystyle \sum_{cycl}\left(\sqrt[n]{a+\sqrt[n]{a}}+\sqrt[n]{a-\sqrt[n]{a}}\right)\lt 18\right)$
Inequality 101 from the Cyclic Inequalities Marathon $\left(\displaystyle \sum_{cycl}\frac{c^5}{(a+1)(b+1)}\ge\frac{1}{144}\right)$
Sladjan Stankovik's Inequality With Constraint II $(a^4+b^4+c^4+d^2+4abcd\ge 8)$
An Inequality with Constraint in Four Variables $\left(\displaystyle\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b}\ge\frac{1}{8}\right)$
An Inequality with Constraint in Four Variables IV $\left(\displaystyle 27+3(abc+bcd+cda+dab)\ge\sum_{cycl}a^3+54\sqrt{abcd}\right)$
Cyclic Inequality with Square Roots And Absolute Values $\left(\displaystyle \prod_{cycl}\left(\sqrt{a-a^2}+\frac{1}{2\sqrt{2}}|3a-1|\right)\ge\frac{1}{6\sqrt{6}}\prod_{cycl}\left(\sqrt{a}+\frac{1}{\sqrt{3}}\right)\right)$
From Six Variables to Four - It's All the Same $\left(\displaystyle \frac{5}{2}\le a^2+b^2+c^2+d^2\le 5\right)$
Michael Rozenberg's Inequality in Two Variables $(\displaystyle \sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6)$
Dan Sitaru's Cyclic Inequality in Three Variables II $\left(\displaystyle \sum_{cycl}\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}\geq \frac{9}{12-2(ab+bc+ca)}+3\right)$
Dorin Marghidanu's Two-Sided Inequality $\left(\displaystyle \small{64(a+bc)(b+ca)(c+ab)}\le \small{8(1-a^2)(1-b^2)(1-c^2)}\le \small{(1+a)^2(1+b)^2(1+c^2)}\right)$
Problem 6 from Dan Sitaru's Algebraic Phenomenon $(x\sqrt{y+1}+y\sqrt{z+1}+z\sqrt{x+1}\le 2\sqrt{3})$
A Warmup Inequality from Vasile Cirtoaje $\left(a^4b^4+b^4c^4+c^4a^4\le 3\right)$
An Extension of the AM-GM Inequality $\left(x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{99}x_{100} \le \frac{1}{4}\right)$
- An Extension of the AM-GM Inequality: A second look $\left(x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + \ldots + x_{n-1}x_{n} \le \frac{a^2}{4}\right)$
Distance Inequality $\left(a^2+b^2+c^2\le\displaystyle\frac{9}{2}\right)$
Kunihiko Chikaya's Inequality with a Constraint $\left(4a^3+9b^3+36c^3\ge 1\right)$ An Inequality with Five Variables, Only Three Cyclic $\left(\displaystyle \left(a+\frac{b}{c}\right)^4+ \left(a+\frac{b}{d}\right)^4+ \left(a+\frac{b}{e}\right)^4\ge 3(a+3b)^4\right)$
Second Pair of Twin Inequalities: Twin 1 $\left(\displaystyle \prod_{i=1}^n\left(\frac{1}{a_i^2}-1\right)\ge (n^2-1)^n\right)$
Second Pair of Twin Inequalities: Twin 2 $\left(\displaystyle \prod_{i=1}^n\left(\frac{1}{a_i}+1\right)\ge (n+1)^{n}\right)$
Cyclic Inequality In Three Variables from the 2018 Romanian Olympiad, Grade 9 $\left(\displaystyle \frac{a-1}{b+1}+\frac{b-1}{c+1}+\frac{c-1}{a+1}\ge 0\right)$
Dan Sitaru's Cyclic Inequality in Three Variables IX $\left(\displaystyle \sum_{cycl}\sqrt{(x+y+1)(y+z+1)}\le 6+\sum_{cycl}\frac{x^3+y^3}{x^2+y^2}\right)$
Vasile Cirtoaje's Cyclic Inequality with Three Variables $\left(\displaystyle \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge 2\right)$ Leo Giugiuc and Vasile Cirtoaje's Cyclic Inequality $\left(\displaystyle \sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}+\sqrt{\frac{d}{1-d}}\ge 2\right)$

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