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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.