Fie \( z_1,z_2,z_3\in\mathcal C^* \).Notam \( z=\bar z_1 z_2+\bar z_2 z_3+ \bar z_3 z_1 \), unde \( z\in\mathcal C^* \).Sa se arate ca:
\( \left|z_1 \right|^2+\left|z_2 \right|^2+\left|z_3 \right|^2 \ge Re(z)+\sqrt 3 Im(z) \)
O alta inegalitate cu numere complexe
Moderators: Filip Chindea, Andrei Velicu, Radu Titiu
O alta inegalitate cu numere complexe
Vrajitoarea Andrei
Fie \( z_k=r_k(cosa_k+isina_k) \) pentru \( k=1,2,3 \).
Inegalitatea se rescrie sub forma echivalenta:
\( 2r_1^2+2r_2^2+2r_3^2 \ge \sum (r_1r_2cosa_1cosa_2+r_1r_2sina_1sina_2)+\sqrt{3}\sum(r_1r_2cosa_1sina_2-r_1r_2sina_1cosa_2)= \)
\( =\sum r_1r_2[cos(a_2-a_1)+\sqrt{3}sin(a_2-a_1)]=2\sum r_1r_2sin(a_2-a_1+\frac {\pi} {6} \). In continuare se folosesc inegalitatile \( 1 \ge sin {\alpha} \) si \( a^2+b^2+c^2 \ge ab+bc+ca \).
Inegalitatea se rescrie sub forma echivalenta:
\( 2r_1^2+2r_2^2+2r_3^2 \ge \sum (r_1r_2cosa_1cosa_2+r_1r_2sina_1sina_2)+\sqrt{3}\sum(r_1r_2cosa_1sina_2-r_1r_2sina_1cosa_2)= \)
\( =\sum r_1r_2[cos(a_2-a_1)+\sqrt{3}sin(a_2-a_1)]=2\sum r_1r_2sin(a_2-a_1+\frac {\pi} {6} \). In continuare se folosesc inegalitatile \( 1 \ge sin {\alpha} \) si \( a^2+b^2+c^2 \ge ab+bc+ca \).
Intradevar, solutia ta cu forma trigonometica este eleganta 
.Acum solutia mea :
Avem arhicunoscuta inegalitate in triunghi a lui H. Hadwiger:
\( a^2+b^2+c^2 \ge 4 sqrt 3 S \)
Consideram acum un plan convex cu \( A(z_1),B(z_2),C(z_3) \) unde avem ca:
\( a=\left|z_2 - z_3 \right| \), \( b=\left|z_3 - z_1 \right| \) , \( c=\left|z_1 - z_2 \right| \) si
\( S=\frac{1}{2} Im(\bar z_1 z_2+ \bar z_2 z_3 +\bar z_3 z_1)=\frac{1}{2} Im(z) \)
Avem ca:
\( \sum \left|z_2-z_3 \right|^2=\sum (z_2-z_3)(\bar z_2-\bar z_3)=\sum (\left|z_2\right|^2+\left|z_3\right|^2-\bar z_2 z_3 -\bar z_3 z_2)=2(\left|z_1\right|^2+\left|z_2\right|^2+\left|z_3\right|^2) - (z+\bar z) \)
Inegalitatea se rescrie sub forma echivalenta:
\( \left|z_1\right|^2+\left|z_2\right|^2+\left|z_3\right|^2 \ge \frac{z+\bar z}{2}+sqrt 3 Im(z) \)
Cum \( Re(z)=\frac{z+\bar z}{2} \) rezulta inegalitatea din problema
.Acum solutia mea :Avem arhicunoscuta inegalitate in triunghi a lui H. Hadwiger:
\( a^2+b^2+c^2 \ge 4 sqrt 3 S \)
Consideram acum un plan convex cu \( A(z_1),B(z_2),C(z_3) \) unde avem ca:
\( a=\left|z_2 - z_3 \right| \), \( b=\left|z_3 - z_1 \right| \) , \( c=\left|z_1 - z_2 \right| \) si
\( S=\frac{1}{2} Im(\bar z_1 z_2+ \bar z_2 z_3 +\bar z_3 z_1)=\frac{1}{2} Im(z) \)
Avem ca:
\( \sum \left|z_2-z_3 \right|^2=\sum (z_2-z_3)(\bar z_2-\bar z_3)=\sum (\left|z_2\right|^2+\left|z_3\right|^2-\bar z_2 z_3 -\bar z_3 z_2)=2(\left|z_1\right|^2+\left|z_2\right|^2+\left|z_3\right|^2) - (z+\bar z) \)
Inegalitatea se rescrie sub forma echivalenta:
\( \left|z_1\right|^2+\left|z_2\right|^2+\left|z_3\right|^2 \ge \frac{z+\bar z}{2}+sqrt 3 Im(z) \)
Cum \( Re(z)=\frac{z+\bar z}{2} \) rezulta inegalitatea din problema
Vrajitoarea Andrei