IMO LL Inequality problem

[sirio11]

This was proposed by USA for IMO 1979, it did not make the short list (LL stands for long list), but I think it's a good inequalities problem.

If a_1,a_2,...,a_n denote the length of the sides of an arbitrary n-gon, prove that:

[n/(n-1)] <= [a_1/(S - a_1)] + [a_2/(S - a_2)] +..... + [a_n/(S - a_n)] <= 2


where S = a_1 + a_2 +......+ a_n

[TomCowley]

<font color="white">Assume the numbers are in decreasing order a1&gt;=a2&gt;=a3...

Probably a prettier solution, and this is a little hand-wavy, but: For a given S, when all but 2 numbers are held constant, varying the other two numbers has the following property (shown easily for a1&gt;a2 and 0&lt;x&lt;a2 and 0&lt;x&lt;(a1-a2) by making the expressions a1+x/(S-(a1+x)) and a2-x/(S-(a2-x)) (or a1-x and a2+x)), increasing the larger number increases the value of the expression, and increasing the smaller number decreases the value of the expression. Given a constant S, minimizing the function requires that no number be bigger than any other number (otherwise it could be reduced by increasing the smaller number and decreasing the larger number), therefore all numbers must be the same. When this is true, a1=S/n, and the expression is equal to n*(S/n)/(S-S/n)=n/n-1.

Maximizing the function is similar. The biggest number must always be increased, but subject to the limit that a1&lt;=a2+a3+..+an because these are sides of an n-gon (extension of the triangle inequality a1&lt;=a2+a3). So increasing a1 up to S/2 at the expense of the other numbers (in reverse a_n, a_n-1, etc. order) will increase the value of function. Next, holding a1=S/2 constant, increasing a2 to S/2 at the expense of the remaining numbers will increase the function. So we get Approaching S/2, approaching S/2, approaching 0, approaching 0.. as the numbers. The function is then maximized at approaching (S/2)/(S-S/2) + (S/2)/(S-S/2) + 0/S +0/S.... = 1+1+0... = 2.</font>

[blah_blah]

The left inequality has nothing to do with the n-gon condition, and here is a reasonably nice proof. By homogeneity we may assume that the sum of the a_i's is 1 (that is, S=1). By Jensen's inequality, since the function 1/x is convex, taking the a_i's as weights yields

\sum_i a_i/(S-a_i) \geq (\sum_{i\not=j} a_ia_j)^{-1}

Finally, comparing this with the 'obvious' inequality

(n-1)/n = (n-1)/n * S^2 \geq (\sum_{i\not=j} a_ia_j)

(this can easily be written as a sum of squares) yields the LHS.

We can also get this inequality in the homogeneous form by using Cauchy Schwarz;

\sum_i a_i^2/(a_i(S-a_i)) \geq S^2/(\sum_{i\not=j} a_ia_j)

but I really view Cauchy Schwarz as a convexity technique anyways. Of course we can use rearrangement or some other similar tool on the left hand side.

For the other part, let a_1 be the longest side.

Then
\sum_i a_i/(S-a_i) = a_1/(S-a_1) + \sum_{i&gt;1} a_i/(S-a_i)
\leq 1 + \sum_{i&gt;1} a_i/(S-a_1)
= 1+1
=2
where the first inequality follows from the n-gon inequality (a_1 \leq \sum_{i&gt;1} a_i) and from a_1\geq a_i for i&gt;1.

[sirio11]

[ QUOTE ]
The left inequality has nothing to do with the n-gon condition

[/ QUOTE ]

Ya, I noticed when I solved it. I have a solution for the LHS without Jensen's, just using the harmonic and arithmetic means

[blah_blah]

it isn't like jensen's is a particularly deep technique; my post shows that it is exactly equivalent to using Cauchy Schwarz in the obvious way.

[sirio11]

Sum (S/S-a_i) = Sum (1 + a_i/S-a_i) = n + Sum (a_i/S-a_i)

Now the harmonic mean &lt;= arithmetic mean, then:

[ n / Sum (1/(1/S-a_i))] &lt;= [Sum (1/S-a_i)]/n

therefore n^2/Sum(S-a_i) &lt;= Sum (1/S-a_i)

n^2/S(n-1) &lt;= Sum (1/S-a_i) ; n^2/n-1 &lt;= Sum (S/S-a_i) = n + Sum (a_i/S-a_i)

[n^2/n-1]-n &lt;= Sum (a_i/S-a_i) therefore n/(n-1) &lt;= Sum (a_i/S-a_i)

qed

BTW nice proof of the RHS blah.

Nice proof also Tom, I see you use the maximal and minimal approach a lot. Sometimes it's hard to use it (or just hard to explain), but in this problem it's just fine.

[jay_shark]

Jensen's inequality is rarely used in IMO problems and is more reserved for Putnam problems; however , the solution was nice .

I think Tom's solution is fairly elementary and probably intended as one of the solutions they were looking for .

[blah_blah]

Jensen's is well-known to all competitors these days, and, as I said, my application of it is not particularly deep since Cauchy Schwarz gives us

\sum_i a_i^2/(a_i(S-a_i)) \geq S^2/(\sum_{i\not=j} a_ia_j)

which is exactly the same as what my application of Jensen's yields. My solution, at least with Cauchy Schwarz in place of Jensen's, is shorter, more rigorous, and almost certainly 'isomorphic' to the official solution.

[sirio11]

[ QUOTE ]
Jensen's inequality is rarely used in IMO problems and is more reserved for Putnam problems

[/ QUOTE ]

I'd say it would be pretty rare if a IMO participant didn't know Jensen's. I think it's basic inequalities knowledge for the IMO these days.