I have a challenging question today. It looks easy but I have no idea how to prove it, so I need your help. The question is:
Prove that$$\arctan(|x-y|)\leqslant\arctan(|x|)+\arctan(|y|).$$
We have: $$y \geq 0 \implies \arctan(x-y) ≤ \arctan(x)+\arctan(y) \tag{1}$$ $$y \leq 0\implies \arctan(x-y) ≥ \arctan(x)+\arctan(y) \tag{2}$$ $$|\arctan(x)| = \arctan(|x|) \tag{3}$$ It is easy to prove those three formulas.
Let $$\text{Let }\, f(a)=\arctan (a+b) -\arctan (a) \; \text{and} \; a,b >0. \tag{i} $$
$$\text{Then }\, f'(a)=\frac 1 {1+(a+b)^{2}} -\frac 1 {1+a^{2}} <0. \tag{ii}$$
Hence $f$ is a decreasing function with
$$f(0)=\arctan (b),\tag{iii}$$
$$f(a) \leq \arctan \,b \quad \quad a,b \in \Bbb{R}.\tag{iv}$$
Thus $$\arctan (a+b) \leq \arctan (a) +\arctan (b).\tag{v}$$
Now use the fact that $\arctan $ is an increasing function to see that
$$\arctan |x-y| \leq \arctan (|x|+|y|) \leq \arctan |x| +\arctan |y|\tag{vi}.$$