I found out that I'm doing BMO2. I've got some past papers, and struggled over the questions.

Can anyone help me with this inequality:

Given that p+q+r=1, and that p, q, r are all non-negative, prove that
7(pq+qr+rp) < = 2+9pqr

Any help appreciated, thanks

Peter

Clue: try and make both sides of the required inequality homogeneous. Then expand out and see if you can prove the result using AM-GM.

Sorry, I don't quite understand what you mean.

It turns out that I've got a solution by factorising and then using AM-GM. Is that what you mean?

I rearranged it to obtain:

(7/9 - p)(7/9 - q)(7/9 - r) < = (4/9)³

Which is proved by application of AM-GM:

(7/9 - p + 7/9 - q + 7/9 - r)/3 > = [(7/9 - p)(7/9 - q)(7/9 - r)]^1/3

Yielding:

(7/3 -1)/3 = 4/9 > = [(7/9 - p)(7/9 - q)(7/9 - r)]^1/3

Is this what you meant? If not, could you show me how you did it, because I'd like to get as much experience with these BMO2 problems as possible.

Also, if anyone has any past papers from '98 backwards could they post them or email them.

Thanks

Peter

Your method is nice - just a small point: how do you know that 7/9 - p, 7/9 - q, 7/9 - r are positive? Anyway, you can probably check the case when they're not positive separately.

The way I was thinking was less elegant but it can be used to solve lots of problems of this type. We want to prove:

7(pq+qr+rp) < = 2+9pqr

by AM-GM. Notice that as (p + q + r) = 1 this problem can also be written:

7(pq + qr + rp)(p + q + r) < = 2(p + q + r)³ + 9pqr (*)

Now this is a homogeneous equation - i.e. each term is of order three. This helps when applying AM-GM as we will now see. If you expand out (*) it becomes:

p² q + p² r + q² r + q² p + r² p + r² q < = 2(p³ + q³ + r³ ) (**)

So that is what we want to prove. But by AM-GM:

p² q < = (p³ + p³ + q³ )/3
p² r < = (p³ + p³ + r³ )/3
q² r < = (q³ + q³ + r³ )/3
q² p < = (q³ + q³ + p³ )/3
r² p < = (r³ + r³ + p³ )/3
r² q < = (r³ + r³ + q³ )/3

Add these six inequalities together and you get (**), which completes the proof.

I remember actually doing this problem in BMO2 two years ago, not knowing AM-GM. I tried to use calculus but it just got horribly messy so I didn't get anywhere. In general it seems that all BMO questions are rigged so that any calculus approach fails miserably, so never be tempted to use it!

Of course, it is not a problem if one of 7/9 - p, 7/9 - q, 7/9 - r is negative. As p + q + r = 1 we know that if one of p,q,r > 7/9 then the other two are < 7/9 so (7/9 - p)(7/9 - q)(7/9 - r) is negative so it is clearly < = (4/3)³ so your solution does work. You would have to explicitly deal with the case where one of p,q,r is > 7/9 though as AM-GM doesn't apply to negative numbers.

In BMOs, it is normally the case that if you can homogenise (i.e. every term is of the same degree) or get symmetry in the inequality, then it should be done and the answer comes out after applying AM-GM or Cauchy-Schwarz.

Anyway, just to add to Michael's point: if you can avoid using calculus in Olympiad question, by all means avoid it. Those questions are designed so that calculus gives very messy solution. For example, in this inequality you will need to use Lagrange's multiplier and you will need to mention 'closed and bounded' a lot. Having said that, if you have no idea how to do the question, calculus could probably tell you when equality holds and you can, hopefully, find a (neat) solution using this information.

Kerwin