<s>[INST] Inequality with complex numbers involving modules Let $z_1,z_2,z_3 \in \Bbb C$ so as $|z_1|=|z_2|=|z_3|=1$ and let $a=|z_1-z_2|, b=|z_2-z_3|, c=|z_3-z_1|$. Show, using algebraic methods, that:
$$ \frac {1} {(a+b-c)^2}+ \frac {1} {(b+c-a)^2}+ \frac {1} {(a+c-b)^2} \ge 1.$$
 [/INST] This solution is a bit geometric. I'll use the fact that $a,b,c$ are the sides of a triangle inscribed in a circle of radius $1$. It's clear that $$a+b+c\le 3\sqrt3.$$ Now call $a+b-c=x$, $a+c-b=y$, $b+c-a=z$. We have that $x,y,z>0$ and $x+y+z=a+b+c$. By Cauchy-Schwarz and AM-HM
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{3}\ge\frac{3^3}{(x+y+z)^2}\ge\frac{3^3}{(3\sqrt3)^2}=1$$
and this is the desired inequality.
</s>