21Using power mean inequality instead of holders - - - -
as i fail to remember holder and minkowski...
\hspace{-16}$Find all Complex no. $\mathbf{z_{1}\;,z_{2}\;,z_{3}\in\mathbf{C}}$ in \\\\\\ $\mathbf{\begin{Vmatrix} \bold{\hspace{-70}z^3_{1}+z^3_{2}+z^3_{3}=24} \\\\ \bold{\left(z_{1}+z_{2}\right).(z_{2}+z_{3}).(z_{3}+z_{1})=64} \\\\ \bold{\mid z_{1}+z_{2}\mid=\mid z_{2}+z_{3}\mid =\mid z_{3}+z_{1}\mid} \end{Vmatrix}}$
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6 Answers
71One obvious solution which comes to mind is (2,0) (2,0) and (2,0).
341We have |z_1+z_2| = 4 \le |z_1|+|z_2|
Adding this to the two other similar inequalities we get |z_1|+|z_2|+|z_3| \ge 6
Now from Holder's inequality we obtain
(1+1+1)(1+1+1)(|z_1|^3+|z_2|^3+|z_3|^3) \ge (|z_1|+|z_2|+|z_3|)^3 \ge 6^3 = 216
|z_1|^3+|z_2|^3+|z_3|^3 \ge 24
But we are given that |z_1|^3+|z_2|^3+|z_3|^3 = 24
This means that |z_1|=|z_2|=|z_3| =2
and further since |z_1+z_2|=|z_1|+|z_2|, this means z1 and z2 are in the same line segment through origin
But since |z_1|=|z_2|, this also means that z_1=z_2
Using the second condition we get z_1=z_2=z_3=2 as the unique solution
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