Fie n numere naturale , nenule si distincte \( a_1,a_2,..a_n \) astfel incat dintre oricare doua unul este divizorul celuilalt. Sa se arate ca :
\( \frac{1}{a_1^2}+\frac{1}{a_2^2}+...+\frac{1}{a_n^2}<\frac{4}{3} \)
M.E.Panaitopol,G.M.
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Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Presupunand ca \( a_1<a_2<...<a_n \) atunci \( a_1|a_2|...|a_n \)
de unde \( a_2\ge 2a_1 \) , \( a_3\ge 2^2a_1 \) , ...,\( a_n\ge 2^{n-1}a_1 \) si de aici concluzia e evidenta.
Adica \( LHS<\frac{1}{a_1^2}(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{n-1}})<\frac{1-\frac{1}{4^n}}{1-\frac{1}{4}}<\frac{3}{4} \)
de unde \( a_2\ge 2a_1 \) , \( a_3\ge 2^2a_1 \) , ...,\( a_n\ge 2^{n-1}a_1 \) si de aici concluzia e evidenta.
Adica \( LHS<\frac{1}{a_1^2}(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{n-1}})<\frac{1-\frac{1}{4^n}}{1-\frac{1}{4}}<\frac{3}{4} \)