Prove Complex Inequality: $(|z_1 + z_2| + |z_1 - z_2|)(|z_1| + |z_2|)>=\sqrt{2}$

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The discussion focuses on proving the complex inequality $$(|z_1 + z_2| + |z_1 - z_2|)(|z_1| + |z_2|) \geq \sqrt{2}$$ using the triangle inequality $$|A| + |B| \geq |A + B|$$ with specific substitutions for $z_1$ and $z_2$. Initial attempts suggest that the inequality can only establish a lower bound of 1, not $\sqrt{2}$. The participants propose substituting $z_1 = e^{i\phi_1}\cos\theta$ and $z_2 = e^{i\phi_2}\sin\theta$ to further explore the inequality.

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Prove that $$(|z_1 + z_2| + |z_1 - z_2|)(|z_1| + |z_2|) >= \sqrt{2}$$
 
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Apply the inequality $$|A|+|B|\geq |A+B|$$ for ##A=z_1+z_2, B=z_1-z_2##

Then apply it again for ##A=z_1+z_2,B=z_2-z_1##.

You ll get two inequalities, add them and it should be straightforward to proceed.

EDIT: Well, using the above suggestion I think you can only prove a lower bound of 1 not ##\sqrt{2}##.
 
Last edited:
The given condition says
z_1=e^{i\phi_1}\cos\theta
z_2=e^{i\phi_2}\sin\theta
How about substituting them in the forlmula?
 

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