# 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

### 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.

[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny