[Search for users] [Overall Top Noters] [List of all Conferences] [Download this site]

Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1996.0. "IMO 95 Q2" by JOBURG::BUCHANAN () Mon Sep 11 1995 21:47

T.RTitleUserPersonal
Name
DateLines
1996.1SolutionEVTSG8::ESANUAu temps pour moiTue Sep 12 1995 13:1243
The inequality to be proved is equivalent to:

	                    3
	--------------------------------------------- <= 2
	      1               1               1
	------------- + ------------- + -------------
	a^3 * (b + c)   b^3 * (c + a)   c^3 * (a + b)

Apply the inequality Harmonic Mean <= Geometric Mean to

	x = a^3 * (b + c)
	y = b^3 * (c + a)
	z = c^3 * (a + b)

i.e.

	          3
	--------------------- <= (x * y * z)^(1/3)
	1 / x + 1 / y + 1 / z

and you get

	                    3
	--------------------------------------------- <=
	      1               1               1
	------------- + ------------- + -------------
	a^3 * (b + c)   b^3 * (c + a)   c^3 * (a + b)

	<= [(a^3 * b^3 * c^3) * (a + b) * (b + c) * (c + a)]^(1/3)

The RHS must be shown <= 2.

As  a * b * c = 1 , this amounts to

	(a + b) * (b + c) * (c + a) <= 8

Using again the hypothesis and the inequality

	x / y + y / x <= 2

this is easily done.

Mihai.
1996.2not convincedJOBURG::BUCHANANTue Sep 12 1995 20:2016
>	(a + b) * (b + c) * (c + a) <= 8
>
>Using again the hypothesis and the inequality
>
>	x / y + y / x <= 2
>
>this is easily done.
    
	Hang on. This isn't true! Suppose that a = 2, b = 1, c = 1/2. Then
    the lhs is 45/4, which is > 8.
    
    	By the way, I should admit that I haven't managed to crack this one
    myself yet...
    
    Cheers,
    Andrew.
1996.3Ooops...EVTSG8::ESANUAu temps pour moiWed Sep 13 1995 08:269
The inequality is the other way round:

	x / y + y / x >= 2

(equivalent with  (x - y)^2 >= 0 )

You see now why I never got on the IMO team? :-)

Mihai.
1996.4Second attemptEVTSG8::ESANUAu temps pour moiTue Sep 19 1995 12:52166
I was foolish to think that a solution to an IMO problem could be based
upon a single idea (namely the inequality  Harmonic Mean <= Geometric Mean
). IMO problems are at least 2-tiered, with at least one vicious idea.

So...


Proposition.  Let  a, b, c  be positive reals such that  abc=1 , and let
p  pe an integer,  p != 0, 1 . Then


	    1           1           1        3
P(p)	--------- + --------- + --------- >= -
	a^p (b+c)   b^p (c+a)   c^p (a+b)    2



Notation. Let  E(p)  denote the LHS of the above inequality.


Notation. We'll use the the notation

	/SIGMA f(a,b,c) = f(a,b,c) + f(b,c,a) + f(c,a,b)

This way P(p) is rewritten as

                          1        3
P(p)	E(p) = /SIGMA --------- >= -
                      a^p (b+c)    2


R0. I have a problem with  P(0)  and  P(1)  (I don't even know whether they
are true), so I'll prove  P(p)  for  p /in Z \ {0,1} . (The problem is in
the indiuctive step below, where I can't take  p = 0 ).


Proof of R0:
------------

We shall use the following inequality:


Lemma. (Ehlers's inequality)
----------------------------
Let  x(1),...,x(n)  be positive reals such that  x(1)...x(n) = 1 . Then

	x(1) + ... + x(n) >= n

the equality being obtained if and only if  x(1) = ... = x(n) = 1


Proof of the lemma: This inequality hardly deserves to carry a name, as it
is a straightforward application of the inequality  Geometric Mean <= <=
Arithmetic Mean . However, it's a good exercise to prove it directly.

q.e.d. (Lemma)


R1.  P(-p) <==> P(p+1)

Proof of R1:
------------

One passes from  P(-p)  to  P(p+1)  and back using the transformation
x |--> 1/x (remember that  abc = 1  so that  (1/a)(1/b)(1/c) = 1  also).

q.e.d. R1.


So it suffices to prove  P(p)  for  P >= 2 .



Initial step ( p = 2 ): P(2)  is equivalent, by R1, to  P(-1) , so let's
------------
prove  P(-1) . We must prove that


	 a     b     c     3
	--- + --- + --- >= -
	b+c   c+a   a+b    2

This is equivalent to

	2(a(c+a)(a+b) + b(b+c)(a+b) + c(b+c)(c+a)) >=
		>= 3(a+b)(b+c)(c+a)

Using the hypothesis, this amounts to

	2a^3 + 2b^3 + 2^3 >= a^2 b + ab^2 + b^2 c + bc^2 + c^2 a + ca^2
i.e.
	a^2 (a-b) + a^2 (a-c) + b^2 (b-a) + b^2 (b-c) + c^2a(c-a) +
		+ c^2 (c-b) >= 0
i.e.
	(a-b)(a^2 - b^2) + (b-c)(b^2 - c^2) + (c-a)(c^2 - a^2) >= 0
i.e.
	(a-b)^2 (a+b) + (b-c)^2 (b+c) + (c-a)^2 (c+a) >= 0

which is obvious for positive  a,b,c .


Here is a more elegant proof of the initial step, using Ehlers's inequality.
Consider the triplets

	a+b  b+c  c+a           c+a  a+b  b+c
	---, ---, ---    and    ---, ---, ---
	b+c  c+a  a+b           b+c  c+a  a+b

We get

	a+b   b+c   c+a                c+a   a+b   b+c
	--- + --- + --- >= 3    and    --- + --- + --- >= 3
	b+c   c+a   a+b                b+c   c+a   a+b

Add them to obtain

	2a+b+c   2b+c+a   2c+a+b
	------ + ------ + ------ >= 6
	  b+c      c+a      a+b

i.e.

	   a     b     c
	2(--- + --- + ---) + 3 >= 6
	  b+c   c+a   a+b

q.e.d. (Initial step)


Inductive step:
---------------

Suppose	 P(p)  is true,  p >= 2 .

                  1                    1
E(p) = /SIGMA --------- = /SIGMA (------------)^p
              a^p (b+c)           a (b+c)(1^p)

                  1                           1
E(p+1) = /SIGMA ------------ = /SIGMA (----------------)^(p+1)
               a^(p+1) (b+c)           a (b+c)(1^(p+1))

We shall apply the following lemma:


Lemma 2 (Jensen's inequality)
-----------------------------

If  0 < r < s , then  (/SIGMA a^s)^(1/s) < (/SIGMA a^r)^(1/r)

unless all the a but one are zero.


Thence, as 0 < 1/(p+1) < 1/p , we have

E(p)^(1/p) < E(p+1)^(1/(p+1))

which implies E(p+1) > E(p)^(1+1/p) > E(p) >= 3/2 by the inductive
hypothesis.

q.e.d. the inductive step.

---

Ouf!
Mihai.