Se considera numerele complexe \( a,b,c \) distincte doua cate doua, astfel incat \( |a|=|b|=|c|=1 \) si \( {|a-b|}^2+{|b-c|}^2+{|c-a|}^2>8 \).
Sa se arate ca \( |(a+b)(b+c)(c+a)| \leq 1 \).
|a|=|b|=|c|=1 , atunci...
Moderators: Filip Chindea, Andrei Velicu, Radu Titiu
-
Marius Dragoi
- Thales
- Posts: 126
- Joined: Thu Jan 31, 2008 5:57 pm
- Location: Bucharest
|a|=|b|=|c|=1 , atunci...
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers
Daca desfacem dupa relatia \( |z|^2 = z \cdot \overline{z} \) , rezulta :
\( {|a-b|}^2+{|b-c|}^2+{|c-a|}^2>8 \Rightarrow 1 > |a+b+c| \) .
Relatia obtinuta ma duce cu gandul la o interpretare geometrica .
Daca avem \( A(a) , B (b) , C(c) \) , triunghiul \( \triangle{ABC} \) este ascutitunghic .
Este cunoscut si faptul ca \( |b+c|=2R \cos{A}=2 \cos{A} \) . Inmultind cu analoagele , ramane de demonstrat ca :
\( \cos{A} \cos{B} \cos{C} \leq \frac{1}{8} \Leftrightarrow \sqrt[3]{\cos{A} \cos{B} \cos{C} } \leq \frac{1}{2} \) .
Dar :
\( \sqrt[3]{\cos{A} \cos{B} \cos{C}} \leq \frac{\cos{A} + \cos{B}+ \cos{C}}{3} \leq \frac{\frac{3}{2}}{3}= \frac{1}{2} \) .
\( {|a-b|}^2+{|b-c|}^2+{|c-a|}^2>8 \Rightarrow 1 > |a+b+c| \) .
Relatia obtinuta ma duce cu gandul la o interpretare geometrica .
Daca avem \( A(a) , B (b) , C(c) \) , triunghiul \( \triangle{ABC} \) este ascutitunghic .
Este cunoscut si faptul ca \( |b+c|=2R \cos{A}=2 \cos{A} \) . Inmultind cu analoagele , ramane de demonstrat ca :
\( \cos{A} \cos{B} \cos{C} \leq \frac{1}{8} \Leftrightarrow \sqrt[3]{\cos{A} \cos{B} \cos{C} } \leq \frac{1}{2} \) .
Dar :
\( \sqrt[3]{\cos{A} \cos{B} \cos{C}} \leq \frac{\cos{A} + \cos{B}+ \cos{C}}{3} \leq \frac{\frac{3}{2}}{3}= \frac{1}{2} \) .