prove that $\displaystyle |z_1|^2 + |z_2|^2 + |z_1 + z_2|^2 \ge \frac{ 2(|a\cdot z_1 + b\cdot z_2|^2)}{(a^2 + b^2)}$ $\displaystyle z_1$ and $\displaystyle z_2$ are complex numbers and a and b are non zero real numbers
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Try $\displaystyle z_1=i,\ z_2=-i,\ a=1,\ b=-1.$
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