Bilkent University Problem of the Month

[jay_shark]

The following problem is this months problem of the month . If you know the solution to this problem you can get credit for it on their website . http://www.fen.bilkent.edu.tr/~cvmath/Problem/0702q.pdf

[jay_shark]

For two positive reals a1,a2 , is this true? if so , then perhaps there is a mathematical induction argument that works .

[1+1/(a1^2+a1^4)][1+1/a2^2+a2^4] >=[1+1/(a1a2+(a1a2)^2)]^2

hmmm...

I'm not sure if 1024 plays an important part in this inequality but I doubt it .

[Enrique]

[ QUOTE ]
For two positive reals a1,a2 , is this true? if so , then perhaps there is a mathematical induction argument that works .

[1+1/(a1^2+a1^4)][1+1/a2^2+a2^4] >=[1+1/(a1a2+(a1a2)^2)]^2

hmmm...

I'm not sure if 1024 plays an important part in this inequality but I doubt it .

[/ QUOTE ]

I tried that too.
I tried using repeated AM-GM and it didn't work. I have stopped trying the problem for about a week, maybe I should give it a shot again.

[jay_shark]

I just looked at it today but so far no luck . Maybe some caffeine will inspire me .

coffee time .

[jay_shark]

I verified that it's true for two numbers .I'll show the details later when I hopefully give out a solution .

For two numbers , it turns out that the numerator on the lhs is always >= to the numerator on the rhs . Also , the denominator on the lhs is always <= than the denominator on the rhs . This shows that the lhs is always >= to the rhs .

You have to expand the expression to see that this is true for two numbers . Then mathematical induction should do the trick .

[jay_shark]

Also , i'm very certain that this works only for powers of 2 . Notice that 1024 is 2^10 .

Now induction should certainly be a lot easier to deal with .

[thylacine]

Suggestion:

Let f(x)=1+1/(x+x^2)

Show that for all positive real a,b

f(a^2)f(b^2) >= [f(ab)]^2

or look at concavity of g(z) = f(e^z)

(I have not checked what I am saying)