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From: XXXakoro@hotpop.comXXX (Andres Koropecki)
Newsgroups: sci.math
Subject: Re: IMO problem 2
Date: 6 Aug 2000 00:25:46 -0400
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On 4 Aug 00 11:28:44, Paul Harvey <pharvey@derwent.co.uk> wrote:
>Does there exist an elegant solution to IMO #2, that is
>Let abc = 1, a,b,c are +ve reals
>(a-1+1/b)(b-1+1/c)(c-1+1/a)<=1

This proof is probably less elegant that P. Montgomery's but it's
a different one (and it can probably be simplified too):

If one or three of a+1/b-1, b+1/c-1, c+1/a-1 are negative, the
inequality trivially holds. If two are negative, say b+1/c<1 and
c+1/a<1, then 2+b+1/a <= b+1/c+c+1/a <2. Hence b+1/a < 0, absurd.
Suppose all the three multiplicands are positive. (**)
In the following, we use the facts that abc=1, a=1/bc and such.
(a+1/b-1)(b+1/c-1)(c+1/a-1) <= 1 iff
[(a+1/b-1)/a][(b+1/c-1)/b][(c+1/a-1)/c] <= 1 iff
(1+c-1/a)(1+a-1/b)(1+b-1/c) <= 1 (***_2) iff
[(1+c-1/a)/c][(1+a-1/b)/a][(1+b-1/c)/b] <= 1 iff
(1+1/c-b)(1+1/a-c)(1+1/b-a) <= 1 (***_1).

Suppose 1/c + 1/a + 1/b - a - b - c <= 0 (*)
Since the multiplicands are positive (by (**)) we can use the
arithmetic-geometric means inequality which implies that
(1+1/c-b)(1+1/a-c)(1+1/b-a) <= [(3 + 1/c + 1/a + 1/b - a - b - c)/3]^2
and we know that 3 + 1/c + 1/a + 1/b - a - b - c <= 3 by (*).
Therefore,
[(3 + 1/c + 1/a + 1/b - a - b - c)/3]^2 <= 1 and (***_1) holds. 

If (*) does not hold, we apply a similar argument to (***_2):
The arithmetic-geometric means inequality impies that
(1+c-1/a)(1+a-1/b)(1+b-1/c) <= [(3 + a + b + c - 1/a - 1/b - 1/c)/3]^3
But we have a + b + c - 1/a - 1/b - 1/c <= 0 so that
(1+c-1/a)(1+a-1/b)(1+b-1/c) <= (3/3)^3 = 1 and (***_2) holds.
Q.E.D.