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.

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 $(1/2,1/2)$.
