An inequality involving arctan of complex argument I have the following conjecture:
\begin{equation}
\text{Re}\left[(1+\text{i}y)\arctan\left(\frac{t}{1+\text{i}y}\right)\right] \ge \arctan(t), \qquad \forall y,t\ge0.
\end{equation}
Which seems to be true numerically. 
Can anyone offer some advice on how to approach proving (or disproving) this?
It originates from a question involving the (complex) Hilbert transform of a symmetric non-increasing probability distribution:
\begin{equation}
h(y) = (1+\text{i}y)\int_{-\infty}^\infty \frac{1}{1 + \text{i}(y-t)}\text{d}G(t)
\end{equation}
and attempting to show $\text{Re}[h(y)]$ takes its minimum at $y=0$.
 A: I start at the beginning, with a symmetric non-increasing density $\varphi$ on $\mathbb R$. The  layer cake representation reduces the matter to $\varphi=\chi_{[-a,a]}$. Therefore, the goal is to show that 
$$\int_{-a}^a \operatorname{Re}\frac{1+iy}{1+i(y-t)}\,dt > \int_{-a}^a \operatorname{Re}\frac{1 }{1-i t}\,dt  \tag1$$
for all $y>0$ and $a>0$. Let's find this real part: 
$$
\operatorname{Re}\frac{1+iy}{1+i(y-t)} = \frac{1+y^2-ty}{1+(y-t)^2}  
\tag2$$
and add the value at $-t$ so that we integrate over $[0,a]$ only: 
$$
 \frac{1+y^2-ty}{1+(y-t)^2}+  \frac{1+y^2+ty}{1+(y+t)^2} = 2\,\frac{1+2y^2+t^2+y^4-y^2t^2}{(1+(y-t)^2)(1+(y+t)^2)}
\tag3$$
Looks bad, but we have to keep going. The goal is to prove that
$$
\int_0^a 2\,\frac{1+2y^2+t^2+y^4-y^2t^2}{(1+(y-t)^2)(1+(y+t)^2)}\,dt > 
\int_0^a \frac{2}{1+t^2}\,dt
\tag4$$
Take the difference of two sides, and it simplifies! 
$$
\int_0^a \frac{2 y^2 t^2 (3+y^2-t^2)}{(1+t^2)(1+(y-t)^2)(1+(y+t)^2)}\,dt > 0
\tag5$$
Sweet. The denominator is always positive. The numerator changes sign only once, from plus to minus. 
Therefore, as a function of $a$, the integral in (5)  increases from $0$ (at $a=0$) to some positive 
value, and then decreases to its limit as $a\to\infty$. It turns out that the  limit as $a\to\infty$ is $0$: 
$$
\int_0^\infty \frac{2 y^2 t^2 (3+y^2-t^2)}{(1+t^2)(1+(y-t)^2)(1+(y+t)^2)}\,dt = 0
\tag6$$
which completes the proof. 
Well, I ought to prove (6) instead of trusting my computer. Let
$$
R(z) = \frac{2 y^2 z^2 (3+y^2-z^2)}{(1+z^2)(1+(y-z)^2)(1+(y+z)^2)},\quad z\in\mathbb C
\tag7$$
This is an even rational function which is $O(|z|^{-2})$ at infinity. Hence, $\int_0^\infty R(z)\,dz$
is $\pi i$ times the sum of residues of $R$ in the upper halfplane. For computing the 
residues, it makes sense to go back to where $R$ came from: 
$$
R(z) =   \frac{1+y^2-zy}{1+(y-z)^2}+    \frac{1+y^2+zy}{1+(y+z)^2} -\frac{2}{1+z^2}
\tag8$$
From (8) we easily get 
$$
\begin{split}
\operatorname{res}\limits_{z=i} R(z) &= -\frac{2}{2i} = i \\ 
\operatorname{res}\limits_{z=i-y} R(z) &= \frac{1+y^2+(i-y)y}{2(i-y)} =\frac{1+iy}{2(i-y)} =-\frac{i}{2}\\ 
\operatorname{res}\limits_{z=i+y} R(z) &= \frac{1+y^2-(i+y)y}{2(i+y)} = \frac{1-iy}{2(i+y)}= -\frac{i}{2}  
\end{split} \tag{9}$$
which indeed sum to $0$.  
