Question

Fie a si b doua numere reale, a < b. Demonstrati ca:
$a\ \textless \ \frac {3a+b}{4} \ \textless \ \frac{a+b}{2} \ \textless \ \frac{a+3b}{4} \ \textless \ b$

Answer 1

1) a < \frac{3a+b}4}  <=> 4a<3a+b <=> a<b (A)
2)  \frac{3a+b}4}  <  \frac{a+b}2}  <=> 3a+b<2a+2b <=> a<b (A)
Celelate 2 inegalitati se demonstreaza la fel.