Given $a,b,u \in \mathbb{C}$ such $u^2=ab$

Question

Given $a,b,u \in \mathbb{C}$ such $u^2=ab$. Prove that $$|a|+|b|=|\frac{a+b}{2}-u|+|\frac{a+b}{2}+u|$$ I noticed only that $|\frac{a+b}{2}-u|+|\frac{a+b}{2}+u| \le |a| + |b| + 2|u|$.

Answer 1

There is nice geometric interpretation of this equation. First we notice that $a,u$ and $b $ are in geometric progression. So there is $x\in\mathbb{C}$ such that $ u=ax$ and $b=ax^2$, so we have to prove:

$$ 2+2|x|^2 = |1-x|^2+|1+x|^2$$ which is parallelogram identity for pa parallelogram $ABCD$ with vertices at $A=0$, $B=1$, $C=1+x$ and $D=x$.

P.I. for parallelogram with sides $a,b$ and diagonals $e,f$ is $$2a^2+2b^2=e^2+f^2$$