Undergrad 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|) ≥ √2. Participants suggest applying the triangle inequality to the expressions A = z_1 + z_2 and B = z_1 - z_2, leading to two derived inequalities. However, it is noted that this approach only establishes a lower bound of 1, not √2. An alternative method proposed involves substituting z_1 and z_2 in terms of their polar forms. The conversation emphasizes the need for a more effective strategy to achieve the desired result.
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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}##.
 
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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?
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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