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Fie \( a, b, c \) trei numere complexe, astfel incat

\( a|bc|+b|ca|+c|ab|=0 \).

Demonstrati ca

\( |(a-b)(b-c)(c-a)|\geq 3\sqrt{3}|abc| \).

Bogdan Enescu

Fie \( x=a|bc| \) \( y=b|ac| \) \( z=c|ab| \)
\( x+y+z=0 \) rezulta \( y=\epsilon z \) \( x=\epsilon y \)
\( \frac{x}{y}=\frac{a|b|}{b|a|} \) rezulta \( \frac{a}{b}=\frac{|a|}{|b|}\epsilon \)
Analog \( \frac{b}{c}=\frac{|b|}{|c|}\epsilon \) si deci \( \frac{a}{c}=\frac{|a|}{|c|}\epsilon^2 \)
Daca \( a=0 \) sau \( b=0 \) sau \( c=0 \) avem evident concluzia.
Daca a,b,c nenule impartim cu \( |abc| \) in relatia ceruta si obtinem ca avem de demonstrat ca:
\( |\frac{a}{c}-\frac{b}{c}||\frac{b}{a}-\frac{c}{a}||\frac{c}{b}-\frac{a}{b}|\geq 3\sqrt{3} \)
deci ca:
\( |\frac{|a|}{|c|}\epsilon^2-\frac{|b|}{|c|}\epsilon||\frac{|b|}{|a|}\epsilon^2-\frac{|c|}{|a|}\epsilon||\frac{|c|}{|b|}\epsilon^2-\frac{|a|}{|b|}\epsilon|\geq 3\sqrt{3} \)
de unde, dupa un mic calcul:
\( |||\frac{c}{a}|+|\frac{a}{b}|+|\frac{b}{c}|-\epsilon(|\frac{c}{b}|+|\frac{b}{a}|+|\frac{a}{c}|)\geq 3\sqrt 3 \)
Fie \( u=|\frac{c}{a}|+|\frac{a}{b}|+|\frac{b}{c}| \) si v=\( |\frac{c}{b}|+|\frac{b}{a}|+|\frac{a}{c}| \)
\( \epsilon=-\frac{1}{2}\pm\frac{\sqrt{3}}{2} \)
\( |u-\epsilon v|=\sqrt{(u+\frac{v}{2})^2+\frac{3}{4}v^2}=\sqrt{u^2+v^2+uv} \)
Avem de demonstrat deci ca
\( (|\frac{c}{a}|+|\frac{a}{b}|+|\frac{b}{c})^2+(|\frac{c}{b}|+|\frac{b}{a}|+|\frac{a}{c})^2+(|\frac{c}{b}|+|\frac{b}{a}|+|\frac{a}{c}|)(|\frac{c}{a}|+|\frac{a}{b}|+|\frac{b}{c})\geq 27 \),
care este echivalent cu
\( (|\frac{c^2}{a^2}|+|\frac{a^2}{c^2}|)+(|\frac{a^2}{b^2}|+|\frac{b^2}{a^2}|)+(|\frac{b^2}{c^2}|+|\frac{c^2}{b^2}|)+2(|\frac{c}{b}|+|\frac{b}{c}|)+2(|\frac{b}{a}|+|\frac{a}{b}|)+2(|\frac{a}{c}|+|\frac{c}{a}|)+ \)
\( +(|\frac{c^2}{ab}|+|\frac{ab}{c^2}|)+(|\frac{b^2}{ac}|+|\frac{ac}{b^2}|)+(|\frac{a^2}{bc}|+|\frac{bc}{a^2}|)+3\geq 27 \),
lucru evident, fiecare paranteza fiind \( \geq 2 \).