Hlawka's inequality proof in $\mathbb{C}$ with integrals Does anybody know a proof of Hlawka’s inequality using integrals?
You can find the inequality here
http://www.cut-the-knot.org/arithmetic/algebra/Hlawka.shtml
Usually it’s tackled with the triangle inequality and some algebraic manipulation. I vaguely remember seeing a calculus based proof, but I don’t remember the details.
This is the inequality: a, b and c are complex numbers, prove
$|a| + |b| + |c| + |a+b+c| \geq |a+b| + |a+c| + |b+c|$
 A: @Noam D. Elkies posted one solution for another inequality involving absolute values
by using integral: https://mathoverflow.net/questions/167685/absolute-value-inequality-for-complex-numbers/167741
I tried to write down the details and put it here.
First, let us prove the inequality for real numbers $a, b, c$. 
WLOG, assume that $a + b \ge 0$ and $a + c \ge 0$.
If $b + c\ge 0$, then $\mathrm{RHS} = a + b + c + (a+b+c)$.
If $b + c < 0$, then $\mathrm{RHS} = a + (-b) + (-c) + (a+b+c)$.
The desired result follows.
Second, let us prove the inequality for complex numbers $a, b, c$.
From the inequality for real number $a, b, c$, by using the identity
$$|z| = \frac{1}{4}\int_0^{2\pi} |\mathrm{Re}(\mathrm{e}^{\mathrm{i}\theta} z)| \mathrm{d} \theta, \tag{1}$$
the desired result follows (note: $\mathrm{Re}(\mathrm{e}^{\mathrm{i}\theta} z)$ is real).
Proof of the identity in (1): Let $z = r\mathrm{e}^{\mathrm{\phi}}$ with $r \ge 0$.
We have
$$\frac{1}{4}\int_0^{2\pi} |\mathrm{Re}(\mathrm{e}^{\mathrm{i}\theta} z)| \mathrm{d} \theta
= \frac{1}{4} r \int_0^{2\pi} |\cos (\theta + \phi)| \mathrm{d} \theta
= \frac{1}{4} r \cdot 2\int_{-\pi/2}^{\pi/2} \cos x \mathrm{d} x = r.$$
$\phantom{2}$
Remark: I remember another example of using integral to prove inequality involving absolute values (many years ago).
Let $z_1, z_2, \cdots, z_n$ be complex numbers with $|z_1|+|z_2|+\cdots +|z_n|=1$. Denote $[1..n] = \{1, 2, \cdots, n\}$.
Prove that 
$$\max_{J \subseteq [1..n]} \Big|\sum_{i\in J} z_i\Big| \ge \frac{1}{\pi}.$$
Proof: Consider
\begin{align}
\int_0^{2\pi} \sum_{k=1}^n \max\{\mathrm{Re}(z_ke^{-\mathrm{i} t}), 0\} \mathrm{d} t
= \sum_{k=1}^n |z_k| \int_{-\pi/2}^{\pi/2} \cos x \mathrm{d} x = \sum_{k=1}^n 2|z_k| = 2.
\end{align}
Thus, there exists $t_0 \in [0, 2\pi]$ such that
$$\sum_{k=1}^n \max\{\mathrm{Re}(z_ke^{-\mathrm{i} t_0}), 0\} \cdot 2\pi \ge 2. \tag{2}$$
Let $S = \{k \in [1..n] : \ \mathrm{Re}(z_ke^{-\mathrm{i} t_0}) > 0\}$. From (2), we have
$\sum_{k \in S} \mathrm{Re}(z_ke^{-\mathrm{i} t_0}) \ge \frac{1}{\pi}$ or
$$\mathrm{Re} \Big(e^{-\mathrm{i} t_0} \sum_{k \in S} z_k\Big) \ge \frac{1}{\pi}. $$
Then, by using $|u| \ge \mathrm{Re}(u)$, we have
$$\Big|\sum_{k \in S} z_k\Big | \ge \frac{1}{\pi}.$$
We are done.
