IMO 2012 Problem 2 - Solution

Let’s play with the 2nd problem of the International Mathematics Olympiad (IMO) 2012.

Problem

For n >= 3 where n is an integer and A2, ..., An are positive real numbers and A2 . A3 . ... . An = 1, prove the following:

(1 + A2)^2 . (1 + A3)^3 . (1 + A4)^4 . ... . (1 + An)^n > n^n.

Solution

For this problem, we’re going to leverage AM-GM inequality which states (X1 + X2 + ... + Xn) / n >= n-root(X1 . X2 . X3 ... Xn).

Starting with a small step, we’ll see what the form of AM-GM is for (1 + A2). Based on the AM-GM formula, the inequality form of (1 + A2) will be (1 + A2) / 2 >= sqrt(A2).

However, the trick lies on how we express 1 within (1 + An). Precisely, for 2 <= m <= n, we can rewrite (1 + Am) as ((1 / (m-1)) * (m-1) + Am). Note that we replace 1 with 1 / (m-1) which repeated m-1 times.

With the above in mind, let’s apply the AM-GM form for (1 + Am) where 2 <= m <= n.

[1 / (m-1) + 1 / (m-1) + ... + Am] / m >= m-root(Am / (m-1)^m-1)

Move the denominator m from LHS to RHS which yields the following.

[1 / (m-1) + 1 / (m-1) + ... + Am] >= m . m-root(Am / (m-1)^m-1)

Take the power of m for both LHS and RHS.

[1 / (m-1) + 1 / (m-1) + ... + Am]^m >= m^m . (Am / (m-1)^m-1)

From the last inequation above, we could rewrite it as [1 + Am]^m >= m^m . (Am / (m-1)^m-1).

Next, multiply the LHS with (1 + Am)^m where m starts from 2 until n such as the following.

(1+A2)^2 . (1+A3)^3 . (1+A4)^4 . ... . (1+An)^n

The same thing also applies to the RHS. Note that the inequality sign doesn’t change.

2^2 . (A2) . 3^3 . (A3 / 2^2) . 4^4 . (A4 / 3^3) . ... . n^n . (An / (n-1)^n-1)

From the RHS, it’s clear that we can eliminate 2^2, 3^3, 4^4, ..., (n-1)^n-1 and the remaining terms are A2 . A3 . A4 . ... . An . n^n.

Since we know from the problem statement that A2 . A3 . A4 . ... . An equals to 1, then the RHS has only n^n as its term. In other words, we got (1 + A2)^2 . (1 + A3)^3 . (1 + A4)^4 . ... . (1 + An)^n >= n^n.

However, our proof has not completed yet since the inequality sign is still >= instead of >.

Note that equality holds when (1 + Am)^m = m^m . (Am / (m-1)^m-1) for 2 <= m <= n.

From the AM-GM form of (1 + Am), we know that the equality holds when Am equals to 1 / (m-1).

Take a look at the following.

[1 / (m-1) + 1 / (m-1) + ... + Am] / m >= m-root(Am / (m-1)^m-1)

Both LHS and RHS will be equal when Am = 1 / (m-1) as shown below.

1 / (m-1) = m-root(1 / (m-1)^m)
1 / (m-1) = 1 / (m-1)

However, when Am equals to 1 / (m-1), the expression A2 . A3 . A4 . ... . An will return 1 . 1 / 2 . 1 / 3 . ... . 1 / (n-1) or 1 / (n-1)! which cannot be 1. This happens because n >= 3 according to the problem statement.

Therefore, equality cannot hold and we can discard the equal sign from >=. Finally, we end up with the following.

`(1 + A2)^2 . (1 + A3)^3 . (1 + A4)^4 . ... . (1 + An)^n > n^n`

Q.E.D