Two Cyclic Inequalities

Introduction

Two problems discussed below have a curious story to tell. The first was proposed by the Moscow mathematician D. P. Mavlo and published in Matematica, No. 4, 1987. It was intended to be a difficult problem whose solution relied on calculus. But a short elementary solution has been submitted by a Bulgarian boy Zhivko Georgiev, a ninth-grader at the time. Zhivko's solution is given below.

The second problem has been published next year in journal Kvant (No. 11-12, 1988, problem M1136). It was dedicated to the centennial of the American Mathematical Society and was not supposed to be easy, but still elementary. The solution to the problem has been published in Kvant, No. 5, 1989. It was observed (Savchev and Andreescu, see the references at the bottom of the page) that the concise solution sounded more like a hint. But another solution can be obtained by simple substitution from the first problem. Thus it must be possible to proceed backwards and obtain another elementary solution to the latter. (Well, not that I did not try...)

Problem 1

Let $a,b,c$ be positive numbers. Prove that

$\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc},$

and the equality occurs if and only if $a=b=c=1.$

Solution 1 to Problem 1

Observe that

\begin{align}\displaystyle \frac{1}{a(1+b)}+\frac{1}{1+abc}&=\frac{1+a+ab+abc}{a(1+b)(1+abc)}\\ &=\frac{1}{1+abc}\bigg[\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}\bigg]. \end{align}

And similarly

\begin{align}\displaystyle \frac{1}{b(1+c)}+\frac{1}{1+abc}&=\frac{1+b+bc+abc}{b(1+c)(1+abc)}\\ &=\frac{1}{1+abc}\bigg[\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}\bigg], \end{align}

\begin{align}\displaystyle \frac{1}{c(1+a)}+\frac{1}{1+abc}&=\frac{1+c+ac+abc}{c(1+a)(1+abc)}\\ &=\frac{1}{1+abc}\bigg[\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}\bigg]. \end{align}

$\displaystyle x=\frac{1+a}{a(1+b)},\space y=\frac{1+b}{b(1+c)},\space z=\frac{1+c}{c(1+a)}.$

With these, after adding up, what we get in the left-hand side

$\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}$

whereas on the right we get

$\displaystyle \frac{1}{1+abc}\left[x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}\right]\ge \frac{6}{1+abc},$

because for any positive $u,$ $\displaystyle u+\frac{1}{u}\ge 2,$ with equality only when $u=1.$

Putting everything together,

$\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc} \ge \frac{6}{1+abc},$

i.e., the inequality we set out to prove.

Problem 2

Prove the inequality

\begin{align} \displaystyle 3+(A+M+S)+\bigg(\frac{1}{A}+\frac{1}{M}+\frac{1}{S}\bigg)&+\bigg(\frac{A}{M}+\frac{M}{S}+\frac{S}{A}\bigg)\\ &\ge\frac{3(A+1)(M+1)(S+1)}{AMS+1} \end{align}

for all positive $A,M,S.$

Solution 1 to Problem 2

First, trivially, $3+(A+M+S)=(1+A)+(1+M)+(1+S).$ Next, observe that

$\displaystyle (1+A)+\frac{1}{M}+\frac{A}{M}=\frac{(1+A)(1+M)}{M},$ etc.

So, the left-hand side transforms into

$\displaystyle\frac{(1+A)(1+M)}{M}+\frac{(1+M)(1+S)}{S}+\frac{(1+S)(1+A)}{A},$

such that after division by $(1+A)(1+M)(1+S)$ the inequality becomes

$\displaystyle\frac{1}{M(1+S)}+\frac{1}{S(1+A)}+\frac{1}{A(1+M)}\ge\frac{3}{1+AMS},$

which is the earlier inequality.

Solution 2 to Problem 2

Here's the solution published in Kvant:

The proof reduces to a rather artificial (but also ingenious!) algebraic manipulation. Multiply the difference of the left-hand and the right-hand sides by $AMS(1+AMS),$ regroup the summands, and you will obtain

$AM(M+1)(SA-1)^{2}+MS(S+1)(AM-1)^{2}+SA(A+1)(MS-1)^{2}$

a non-negative quantity. This proof was found by P. H. Diananda from the University of Singapore.

References

1. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003, pp. 101-103