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Interesting inequality


Prove that \text{Det}(\frac{1}{a_i+a_j})\geq 0, where a_1,...,a_n are positive integers (i.e. a_i \geq 1).

Solution:

To solve this, we use the following: 0 \leq  \int_0^1 (u_1x^{a_1}+u_2x^{a_2}+...+u_nx^{a_n})^2 dx=\sum_{i,j=1}^n \frac{u_iu_j}{a_i+a_j}. Therefore, if A=(\frac{1}{a_i+a_j}) and u=(u_1,...,u_n) \in \mathbb{R}^n then u A u^T\geq 0 for any u \in \mathbb{R}^n (1).

Now take \lambda an eigenvalue for A and x one of its eigenvectors. We know from (1) that xAx^T\geq 0, but xA x^T=\lambda xx^T=\lambda (x_1^2+x_2^2+...+x_n^2). Since the last expression is not negative, we must have \lambda \geq 0.

Therefore, any eigenvalue for A is not negative. Since the determinant of A is the product of its eigenvalues, it follows that Det(A)\geq 0 as required.

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Categories: Higher Algebra
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