Mathematical and Physical Journal
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Problem A. 493. (November 2009)

A. 493. Prove that

$\displaystyle \sum_{i=1}^n\ \sum_{j=1}^n \frac{a_ia_j}{{(p_i+p_j)}^c}\ge 0$

holds for arbitrary reals $\displaystyle a_1,a_2,\ldots,a_n$, and positive numbers $\displaystyle c,p_1,p_2,\ldots,p_n$.

(5 pont)

Deadline expired on December 10, 2009.

Solution.

$\displaystyle 0 \le \int_0^\infty \left( \sum_{i=1}^n a_i e^{-p_it} \right)^2 t^{c-1} \,\mathrm{d}t = \int_0^\infty \left( \sum_{i=1}^n a_i e^{-p_it} \right) \left( \sum_{j=1}^n a_j e^{-p_jt} \right) t^{c-1} \,\mathrm{d}t =$

$\displaystyle = \int_0^\infty \left( \sum_{i=1}^n \sum_{j=1}^n a_ia_j e^{-(p_i+p_j)t} \right) t^{c-1} \,\mathrm{d}t = \sum_{i=1}^n \sum_{j=1}^n a_ia_j \int_0^\infty t^{c-1} e^{-(p_i+p_j)t} \,\mathrm{d}t =$

$\displaystyle = \sum_{i=1}^n \sum_{j=1}^n \frac{a_ia_j}{(p_i+p_j)^c} \int_0^\infty \big((p_i+p_j)t\big)^{c-1} e^{-(p_i+p_j)t} (p_i+p_j)\,\mathrm{d}t =$

$\displaystyle = \sum_{i=1}^n \sum_{j=1}^n \frac{a_ia_j}{(p_i+p_j)^c} \int_0^\infty u^{c-1} e^{-u} \,\mathrm{d}u = \Gamma(c) \cdot \sum_{i=1}^n \sum_{j=1}^n \frac{a_ia_j}{(p_i+p_j)^c}.$

Statistics:

 4 students sent a solution. 5 points: Nagy 235 János. 3 points: 1 student. 1 point: 2 students.

Problems in Mathematics of KöMaL, November 2009