WLOG $a\leq b\leq c\leq d$. Let $b=a+u$, $c=a+u+v$, $d=a+u+v+w$, where $u,v, w$ are non-negative real numbers. Thus $$a+b+c+d=4a+3u+2v+w=0$$$$\Longleftrightarrow$$$$a=\frac{-3u-2v-w}{4}\text{.}$$So \begin{align*} \max (a,b)+\max (c,d)+\max (a,c)+\max (b,d)+\max (a,d)+\max (b,c)=b+2c+3d&=6a+6u+5v+3w\\ &=6\frac{-3u-2v-w}{4}+6u+5v+3w\\ &=\frac{3u+4v+3w}{2}\\ &\geq 0 \end{align*}, which is true.

Jetze wrote: (a) Prove that for all $a, b, c, d \in \mathbb{R}$ with $a + b + c + d = 0$, \[ \max(a, b) + \max(a, c) + \max(a, d) + \max(b, c) + \max(b, d) + \max(c, d) \geqslant 0. \](b) Find the largest non-negative integer $k$ such that it is possible to replace $k$ of the six maxima in this inequality by minima in such a way that the inequality still holds for all $a, b, c, d \in \mathbb{R}$ with $a + b + c + d = 0$. Here's a simpler solution for part (a) without having to bash. Solution 01. WLOG $\max(a,b,c,d) = a$. Note that \[ 4a \ge a + b + c + d = 0 \implies a \ge 0 \]Hence, \begin{align*} \max(a,b) + \max(a,c) + \max(a,d) + \max(b,c) + \max(b,d) + \max(c,d) &= 3a + \max(b,c) + \max(b,d) + \max(c,d) \\ &\ge 3a + b + c + d \\ &= 2a \\ &\ge 0 \end{align*}Solution 2. Note that $\max(x,y) \ge \frac{1}{2} (x + y)$ for any $x,y \in \mathbb{R}$. Applying this to the original inequality, \[ \max(a,b) + \max(c,d) \ge 0, \max(a,c) + \max(b,d) \ge 0, \max(a,d) + \max(b,c) \ge 0 \] We claim that $k = 2$. We will prove that \[\min(a,b) + \min(c,d) + \max(a,c) + \max(a,d) + \max(b,c) + \max(b,d)\ge 0 \]for any $a,b,c,d$ such that $a + b + c + d = 0$. To see this, note that $\max(a,b) = \frac{1}{2} (a + b + |a - b|)$ and $\min(a,b) = \frac{1}{2} (a + b - |a - b|)$ for any $a,b$. It suffices to prove that \[ |a - c| + |a - d| + |b - c| + |b - d| \ge |a - b| + |c - d| \]which is obvious by Triangle Inequality. To prove that $k \le 2$, if $k = 3$, then by PHP, some number appear twice in $\min(x,y)$, let it be $a$. Then let $a = -3000$, $b = c = d = 1000$. This gives us \[ \text{Value} \le -6000 + 4(1000) = -2000 < 0 \]a contradiction

(a). WLOG $a\ge b\ge c\ge d$. Then: $$\text{LHS}=3a+2b+c\ge\frac32a+\frac32b+\frac32c+\frac32d=0.$$