Fie \( a,b,c \in \mathbb{C} \) avand acelasi modul. Demonstrati ca:
\( |\frac{a}{b}-\frac{b}{a}|+|\frac{b}{c}-\frac{c}{b}|+|\frac{c}{a}-\frac{a}{c}| \leq 3\sqrt{3} \)
RMT 2004
Inegalitate cu numere complexe de acelasi modul
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Radu Titiu
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Inegalitate cu numere complexe de acelasi modul
A mathematician is a machine for turning coffee into theorems.
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Theodor Munteanu
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Re: Inegalitate cu numere complexe de acelasi modul
\(
{\rm Fie a = r(\cos A + i\sin A),\ b = r(\cos B + i\sin B),\ c = r(\cos C + i\sin C)}. \\
|\frac{{\rm a}}{{\rm b}} - \frac{{\rm b}}{{\rm c}}|{\rm = |\cos(A - B) + i\sin(A - B) - \cos(B - C) - i\sin(B - C)| = } \\
\sqrt {{\rm (\cos(A - B) - \cos(B - C))}^{\rm 2} + (\sin (A - B) - \sin (B - {\rm C))}^{\rm 2} } = \\
\sqrt {2 - 2\cos ((A - B) - (B - C))} = 2|\sin \frac{1}{2}(A - C)|. \\
{\rm Fie in planul complex M(a), N(b), P(c) si m(}\angle MNP{\rm ) = }\frac{{\rm 1}}{{\rm 2}}|A - C| \\
{\rm sau m(}\angle {\rm PQR) = }\pi {\rm - }\frac{{\rm 1}}{{\rm 2}}|A - C| \Rightarrow 2|\sin \frac{1}{2}(A - C)| = \sin (\angle MNP){\rm si analoagele}{\rm .} \\
{\rm Acum stim ca \sin A + \sin B + \sin C} \le {\rm sin}\frac{{{\rm A + B + C}}}{{\rm 3}} = \frac{{3\sqrt 3 }}{2}{\rm si aplicata in }\Delta MNP \\
{\rm da concluzia}{\rm .}
\)
{\rm Fie a = r(\cos A + i\sin A),\ b = r(\cos B + i\sin B),\ c = r(\cos C + i\sin C)}. \\
|\frac{{\rm a}}{{\rm b}} - \frac{{\rm b}}{{\rm c}}|{\rm = |\cos(A - B) + i\sin(A - B) - \cos(B - C) - i\sin(B - C)| = } \\
\sqrt {{\rm (\cos(A - B) - \cos(B - C))}^{\rm 2} + (\sin (A - B) - \sin (B - {\rm C))}^{\rm 2} } = \\
\sqrt {2 - 2\cos ((A - B) - (B - C))} = 2|\sin \frac{1}{2}(A - C)|. \\
{\rm Fie in planul complex M(a), N(b), P(c) si m(}\angle MNP{\rm ) = }\frac{{\rm 1}}{{\rm 2}}|A - C| \\
{\rm sau m(}\angle {\rm PQR) = }\pi {\rm - }\frac{{\rm 1}}{{\rm 2}}|A - C| \Rightarrow 2|\sin \frac{1}{2}(A - C)| = \sin (\angle MNP){\rm si analoagele}{\rm .} \\
{\rm Acum stim ca \sin A + \sin B + \sin C} \le {\rm sin}\frac{{{\rm A + B + C}}}{{\rm 3}} = \frac{{3\sqrt 3 }}{2}{\rm si aplicata in }\Delta MNP \\
{\rm da concluzia}{\rm .}
\)