TITLE: Showing that for any complex number $z$, either $ |z + 1| \geq 1/\sqrt{2} $ or $ |z^2 + 1| \geq 1 $
QUESTION [6 upvotes]: I was reading complex Number from A to Z
There was an example which asks to proof that for any complex Number $z$ either
$$ |z + 1| \geq 1/\sqrt{2} $$
 or
 $$  |z^2 + 1| \geq 1 $$ 
In the book proof was by contradiction and I had no problem in understanding it, but what rise my problem is how someone can arrive to such a conclusion from nothing.

REPLY [3 votes]: Geometrically, looking at the equality case, the first equation corresponds to a circle centered at $-1$ with radius $1/\sqrt{2}$, whereas the second is the locus of points whose product of the distances from $i$ and $-i$ is equal to $1$. The second locus of points is called the Lemiscate of Bernoulli. The thesis is equivalent to saying that the Bernoulli Lemniscate and the circle do not intersect.
Rotating the frame of reference the lemniscate is given by the equation https://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli:
$(x^2+y^2)^2=2(x^2-y^2) [1]$
and the rotated circle:
$x^2+(y-1)^2=r^2 [2]$
The optimal $r$ is given when the circle and the lemniscate are tangent, i.e. the point of intersections are only 2. Substituting $x^2$ from [2] into [1] and putting the determinant to zero I get $\sqrt{2-\sqrt{2}} \sim 0.76$ as optimal r, which is greater than $1/\sqrt{2}\sim 0.70$ the one proposed and proves that the circle and the leminscate do not intersect.