Proving the Positivity of a Series Sum using Linear Algebra and Induction

There is a quite elegant proof for proving the positivity of the series sum:

(sum_{1 leq i, j leq n} frac{a_i a_j}{i j} sum_{1 leq i, j leq n} a_i a_j int_{0}^{1} x^{ij-1} , dx int_{0}^{1} left(sum_{i1}^{n} a_i x^{i-1}right)^2 , dx geq 0)

This proof is equivalent to showing that the matrix defined by (H_{n times n}: h_{ij} frac{1}{ij}) is positive definite. The same technique is also used to prove the positive definiteness of the Hilbert matrix where (h_{ij} frac{1}{ij - 1}).

Explanation

Let's consider each pair of numbers in the series. Each pair of numbers is part of four terms:

(frac{a_n a_m}{nm}) (frac{a_m a_n}{mn}) (frac{a_n a_n}{nn}) (frac{a_m a_m}{mm})

The first two terms are the same, so the sum of the four terms is:

(2 frac{a_n a_m}{nm})

We can simplify this to:

(frac{2a_n^2}{2n} frac{a_n^2}{n}) and (frac{2a_m^2}{2m} frac{a_m^2}{m})

Thus, the sum of these four terms is:

(2 cdot frac{a_n a_m}{nm} frac{a_n^2}{n} frac{a_m^2}{m})

Simplifying this further, we get:

(frac{8nm a_n a_m}{4nm} frac{a_n^2}{n} frac{a_m^2}{m} frac{2a_n a_m}{1} frac{a_n^2}{n} frac{a_m^2}{m})

This can be separated into two parts:

(frac{n a_n^2}{n} frac{m a_m^2}{m})

Both of these parts factorize into squares:

(frac{n a_n^2}{(n a_n)} frac{m a_m^2}{(m a_m)} (n a_n)(m a_m))

Since (nm) is positive, this is positive, proving that the numerator and thus the whole fraction is positive.

Further Considerations

This is not yet a full answer, but here is a fact that might prove useful. Let's consider the following sum:

(s_n sum_{0 leq i, j leq n} a_i a_j).

We can see that for (n1), (s_1 a_1^2). Now, assume that (s_n left(sum_{i1}^{n} a_iright)^2). We then have:

(s_{n-1} sum_{0 leq i, j leq n-1} a_i a_j sum_{0 leq i, j leq n} a_i a_j - a_{n}^2 - 2 sum_{i1}^{n} a_n a_i left(sum_{i1}^{n} a_iright)^2 - a_{n}^2 - 2a_{n} sum_{i1}^{n} a_i)

We note that (x - a_{n}^2 x^2 - 2xa_{n} - a_{n}^2). Substituting (x sum_{i1}^{n} a_i), we see that:

(s_{n-1} left(sum_{i1}^{n-1} a_iright)^2) hence we have shown that (s_n left(sum_{i1}^{n} a_iright)^2) by induction. Note that (s_n geq 0).

I performed a few numerical experiments and as far as I can tell, the sum above seems to be bounded above by (s_n) and below by (frac{s_n}{2n}), both of which are positive.