<s>[INST] 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.
 [/INST] 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 $(1/2,1/2)$.
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