Help proving complex inequality

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The discussion focuses on proving the inequality |a|^2 + |b|^2 >= |(a+b)/2|^2 for complex numbers a and b. The user attempts to derive the inequality using the triangle inequality, leading to the expression |(a+b)/2|^2 <= |a|^2 / 4 + |b|^2 / 4 + |a||b| / 2. They suggest that proving |a|^2 + |b|^2 is greater than or equal to this derived expression will complete the proof. The conversation remains open for further contributions to solve the inequality. The thread highlights a common challenge in complex analysis involving inequalities.
JerryG
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This may seem trivial, but for some reason I am having trouble with it. For a and b in the complex plane, I am trying to prove the following:

|a|^2+|b|^2 >= |(a+b)/2|^2

I need this for part of a larger proof.
 
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Since no one has answered yet, I'll give it a go.

Starting from the triangle inequality, we get

|a+b| <= |a| + |b|

=>

|a+b|^2 <= (|a| + |b|)^2 = |a|^2 + |b|^2 + 2|a||b|

=>

|(a+b)/2|^2 <= |a|^2 / 4 + |b|^2 / 4 + |a||b| / 2If we can prove that |a|^2 + |b|^2 >= |a|^2 / 4 + |b|^2 / 4 + |a||b| / 2, then we're done!

Can you go from here?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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