If $O$ is a point in square $ABCD$, then $\angle OAB+\angle OBC+\angle OCD+\angle ODA\ge\frac{3\pi}4$ 
$O$ is a point in square $ABCD$. Show that
$$\angle OAB+\angle OBC+\angle OCD+\angle ODA\ge\frac{3\pi}4$$

My solution is complex numbers. Let $-C=A=1$, $-D=B=i$. So
\begin{align*}&\mathrm{Arg}\left(\frac{B-A}{O-A}\cdot\frac{C-B}{O-B}\cdot\frac{D-C}{O-C}\cdot\frac{A-D}{O-D}\right)
\\={}&\mathrm{Arg}\left(\frac4{O^4-1}\right)=-\mathrm{Arg}(O^4-1).
\end{align*}
Here, $O$ is taken in the square, but $O^4$ is hard to manage. Firstly, when $|O|\le\dfrac{\sqrt2}2$, it can be in any direction then $O^4$’s locus is circle centered at $0$ with radius $\dfrac14$. However if $|O|$ is bigger, only some direction is in the square, namely $O$ moves on four arcs.
So my solution gets stuck here. You could actually change a method completely, but I’d like to see if my method would work.
 A: I don't think what follows is an elegant solution.
But it uses only very basic geometry.

Divide the square into $8$ equal parts  as shown. By symmetry, it suffices to prove that the sum of the four angles is at least $135^0$ if $O$ lies within the square $ IECH$ as shown.
There are several subcases:
$(1)$ $O$ lies on the diagonal $IC$:

$\Delta ODA \cong \Delta OBA \implies \alpha + \beta + \gamma + \delta = 180^o \geq 135^o$
$$$$
$(2)$ $O$ lies on $EI$:

$OD=OC$ and $OA=OB \implies \alpha + \beta + \gamma + \delta = 180^o \geq 135^o$
$$$$
$(3)$ $O$ lies on $EC$:

$\delta = 90^o$ and $\alpha \geq 45^o \implies \alpha + \beta + \gamma + \delta \geq 135^o$
$$$$
$(4)$ $O$ lies inside $\Delta ECI$ but not on the boundary:

$\tan \gamma \gt \tan x \implies \gamma \gt x \implies \gamma + \delta \gt 90^0$
Since $\alpha \gt 45^o$, $ \alpha + \beta + \gamma + \delta \geq 135^o$
$$$$
$(5)$ $O$ lies inside $\Delta IHC$:

$\tan \delta \gt \tan y \implies \delta \gt y \implies \alpha + \delta \gt 90^0$
Since $\gamma \gt 45^o$, $ \alpha + \beta + \gamma + \delta \geq 135^o$
$$$$
The remaining $2$ cases where $O$ lies on $CH$ or $IH$ can be dealt with similarly.
A: Let $S$ be the square $A=1,\ B=i, \ C=-1, \ D=-i$ and :
$$f(z):=z^4-1 \ \ \text{and} \ \ g(z):=z^4$$
Have a look at the following picture :

Fig. 1: The interior of the green "drop" is the image by $g$ of (filled) square $S$. It can also be seen as the image by $g$ of square $OFAE$ or triangle $OAB$, with the  advantage that transformation $g$, restricted either to $OFAE$ of to $OAB$ is bijective. Animated version here.
Your analysis has reduced the issue to the analysis of the opposite of the argument of $f(z)=z^4-1$ (I have replaced $O$ by $Z$ in order to enhance its variable status). As explained in the legend of fig. 1, the image of square $S$  by transformation $g$ is the interior of the drop-like curve  (in green) which, in a second step, is translated by $-1$ (interior of the red "drop").
Remark: The parametric equations of line $AB$ being :
$$x(t)=\frac{\cos(t)}{\cos(t)+\sin(t)}, y(t)=\frac{\sin(t)}{\cos(t)+\sin(t)},\tag{1}$$
the parametric equations of the green drop are :
$$x(t)=\frac{\cos(4t)}{(\cos(t)+\sin(t))^4}, y(t)=\frac{\sin(4t)}{(\cos(t)+\sin(t))^4}$$
Explanation : taking a complex number at the power $n$ amounts to take the power $n$ of its modulus and multiply its argument by $n$.
One can obtain the green boundary by only taking the image by $g$ of line segment $AE$ for the upper part of the "drop" and line segment $AF$ for its lower part ; in fact the (interior of the) drop is the image by $g$ of the square $AEOF$ or of triangle $OAB$ (indeed, other squares or triangles can be obtained through a multiplication by some $i^k$, this multiplier vanishing with power $4$). Please note that the image by function $g$ of line segment $OE$ is line segment $OG$ with $G(-\frac14,0)$.
Now we can conclude. Lines $OG$ and $OH$, tangent to the red drop, permits to conclude that:
$$\frac{3\pi}{4} \le \arg(z^4-1) \le \frac{5\pi}{4} $$
as desired.
The proof that this interior angle symmetrical with respect to $x$-axis is $\pi/2$ comes plainly from the fact that it is the image of the $\pi/2$ angle $\angle DAC$ by analytic (therefore conformal = angle preserving) transformation $g$.
Remarks :

*

*Moving point $Z$ (in the animated version) inside triangle $OAB$, the sum of angles remains in the vicinity of $180°$ ; unless point $M$ is very close to $A$, it's impossible to approach limit angles $3 \pi/4 = 135° $ and $5 \pi/4 = 225°$.


*The denominator in (1) could be written $\sqrt{2}\cos(t-\frac{\pi}{4})$.
A: If someone wants to have an entirely algebraic trigonometric solution, not using complex numbers, nor having any geometric insight, below is one:
Let's assume $A=(0,0), B=(0,1), C=(1,1), D=(1,0)$, and $O=(a,b)$, where $0<a,b<1$.
It is easy to see that:
$$\angle OAB=\arctan \frac{a}{b}\\ \angle OBC=\arctan \frac{1-b}{a}\\\angle OCD=\arctan \frac{1-a}{1-b}\\\angle ODA=\arctan \frac{b}{1-a}.\\$$
Now, if $a=b$, or $a+b=1$, with the help of the relation $\arctan x+ \arctan \frac{1}{x}=\frac{\pi}{2}$, we get that:
$$A=\arctan \frac{a}{b}+\arctan \frac{1-b}{a}+\arctan \frac{1-a}{1-b}+\arctan \frac{b}{1-a}=\pi.$$
Hence four cases happen:
$$1>b+a, b>a \\ 1>b+a, a>b\\a+b>1, a>b\\a+b>1, b>a .$$
Assume $1>b+a, b>a$, or $a+b>1, a>b.$ In these cases, $\frac{1-b}{a}>\frac{1-a}{b}$ So,
$$A=\arctan \frac{a}{b}+\arctan \frac{1-b}{a}+\arctan \frac{1-a}{1-b}+\arctan \frac{b}{1-a}>\\\arctan \frac{a}{b}+(\arctan \frac{1-a}{b}+\arctan \frac{b}{1-a})+\arctan \frac{1-a}{1-b}=\\ \frac{\pi}{2}+\arctan \frac{a}{b}+\arctan \frac{1-a}{1-b}.$$
On the other hand, since either $a>b$ or $b>a$, $\arctan \frac{a}{b}+\arctan \frac{1-a}{1-b}> \frac{\pi}{4}$ (note that $\arctan 1= \frac{\pi}{4}$). As a result, $A>\frac{3\pi}{4}.$
If we assume $1>a+b, a>b$ or $a+b>1, b>a$, in these cases, $\frac {a}{b}>\frac{1-b}{1-a}$. So,
$$A=\arctan \frac{a}{b}+\arctan \frac{1-b}{a}+\arctan \frac{1-a}{1-b}+\arctan \frac{b}{1-a}>\\ (\arctan \frac{1-b}{1-a}+\arctan \frac{1-a}{1-b})+\arctan \frac{1-b}{a}+\arctan \frac{b}{1-a}=\\ \frac{\pi}{2}+\arctan \frac{1-b}{a}+\arctan \frac{b}{1-a};$$
since either $a+b>1$ or $1>a+b$, we have $\arctan \frac{1-b}{a}+\arctan \frac{b}{1-a}> \frac{\pi}{4}$. So, $A>\frac{3\pi}{4}.$
We are done.

Fig. 1 : Representation of the sum of the four angles as a surface $z=f(a,b)$ (for $0<a,b<1$). One notices a natural invariance by $k \pi/2$ rotations around point $(1/2,1/2)$.
