Showing that for any complex number $z$, either $ |z + 1| \geq 1/\sqrt{2} $ or $ |z^2 + 1| \geq 1 $ 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.
 A: Let us think of complex plane. I used the program, GeoGebra to draw the figure below. 
Let $A = 0$, $B = -0.5+0.5i$, $C = -0.5 -0.5 i$, and $D=-1$. 
Your first inequality is true if and only if  $z$ is on the circle or outside the circle. 

Thus it is enough to show that if $z$ is in the circle, then that $z$ satisfies the second inequality. 
The circle is between the half-line $\overrightarrow{AB}$ and $\overrightarrow{AC}$. 
The angle of $\overrightarrow{AB}$ is $\frac{3\pi}{4}$, and the angle of 
$\overrightarrow{AC}$ is $\frac{5\pi}{4}$. 
Generally, assume that $z$ is a complex number between these two half-line. 
We know that the angle of $z^2$ is twice of the angle of $z$. 
Thus the angle of $z^2$ is between $\frac{3\pi}{2}$ and $\frac{5\pi}{2}$. 
It means that the real part of $z^2$ is non-negative. 
Thus, the real part of $z^2 +1$ is greater than or equal to $1$. 
Because $|z^2 + 1| \geq |Re(z^2 +1)|$, it follows that $|z^2 + 1| \geq 1$. Q.E.D.

I think that this geometric proof let us be able to make similar problems. 
For example, let the second inequality be the same. And let the first inequality change as follows:
$$ |z+r| \leq \frac{r}{\sqrt{2}} , $$
for any fixed $r>0$. 
Then we can say that every complex number $z$ satisfies the (changed) first or second inequality. 
A: 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. 
