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Problema 1, lista scurta 2009

Posted: Sun Apr 19, 2009 8:33 pm
by alex2008
Fie numerele naturale nenule \( a_1,a_2,...,a_{n+1} \) si \( k \) astfel incat :

\( a_1\le a_2\le ...\le a_{n+1}. \)

Demonstrati inegalitatea :

\( \left\lfloor \frac{a_2-a_1}{k}\right\rfloor+\left\lfloor \frac{a_3-a_2}{k}\right\rfloor+...+\left\lfloor \frac{a_{n+1}-a_{n}}{k}\right\rfloor+n-1\ge \left\lfloor \frac{a_{n+1}-a_1}{k}\right\rfloor \)

Manea Cosmin si Petrica Dragos, Pitesti

Posted: Sun Apr 19, 2009 10:23 pm
by Marius Mainea
Folosim inegalitatea:
\( [x_1+x_2+...+x_n]\le [x_1]+[x_2]+...+[x_n]+n-1 \)