| > (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.
|
| 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.
|