Arad 2001

[alex2008]

Sa se arate , fara a efectua adunarea , ca :\( \frac{1}{4}+\frac{1}{5}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{4\cdot4}+\frac{1}{5\cdot2}<2 \)

[Natalee]

Sa incercam:

Dintre fractiile: \( \ \ \frac{1}{4}; \ \ \frac{1}{2 \ \cdot \ 4}; \ \ \frac{1}{3 \ \cdot \ 4}; \ \ \frac{1}{4 \ \cdot \ 4 } \), valoarea cea mai mare o are \( \ \frac{1}{4} \)

Dintre fractiile: \( \ \ \frac{1}{5}; \ \ \frac{1}{2 \ \cdot \ 5}; \ \ \frac{1}{3 \ \cdot \ 5}; \ \ \frac{1}{5 \ \cdot \ 2} \), valoarea cea mai mare o are \( \ \frac{1}{5} \).

Notam:

\( S \ = \ \frac{1}{4} \ + \ \frac{1}{5} \ + \ \frac{1}{2 \ \cdot \ 4} \ + \ \frac{1}{2 \ \cdot \ 5} \ + \ \frac{1}{3 \ \cdot \ 4} \ + \ \frac{1}{3 \ \cdot \ 5} \ + \ \frac{1}{4 \ \cdot \ 4} \ + \ \frac{1}{5 \ \cdot \ 2} \)

\( S_{_{_{1}}} \ = \ \frac{1}{4} \ + \ \frac{1}{5} \ + \ \frac{1}{4} \ + \ \frac{1}{5} \ + \ \frac{1}{4} \ + \ \frac{1}{5} \ + \ \frac{1}{4} \ + \ \frac{1}{5} \)

In aceasta situatie: \( S \ < \ S_{_{_{1}}} \)

Calculam \( \ S_{_{_{1}}} \):

\( S_{_{_{1}}} \ = \ 4 \ \cdot \ \frac{1}{4} \ + \ 4 \ \cdot \ \frac{1}{5} \ = \ 1 \ + \ \frac{4}{5} \ = \ \frac{9}{5} \ < \ 2; \ \ \ cum \ S \ < \ S_{_{_{1}}} \ = > \ S \ < \ 2 \)