Can the function $\max(x,y,z)$ be expressed in terms of absolute values (not nested)? Q1: Are there some linear forms $a_ix+b_iy+c_iz$ and signs $\sigma_i\in\{1,-1\}$ such that
$$\max(x,y,z)=(a_0x+b_0y+c_0z)+\sum_{i\geq1}\sigma_i|a_ix+b_iy+c_iz|$$
for all $x,y,z\in\mathbb R$ ?
Q2: Same question for $\max(|x|,|y|,|z|)$.

For the analogous questions with only two variables, the answers are
$$\max(x,y)=\frac{x+y}{2}+\left|\frac{x-y}{2}\right|$$
and
$$\max(|x|,|y|)=\left|\frac{x+y}{2}\right|+\left|\frac{x-y}{2}\right|$$
Show that the $\max{ \{ x,y \} }= \frac{x+y+|x-y|}{2}$.
And the two-variable case can be applied to the three-variable case:
$$\max(x,y,z)=\max(\max(x,y),z)$$
$$=\frac{\frac{x+y+|x-y|}{2}+z+\left|\frac{x+y+|x-y|}{2}-z\right|}{2}$$
$$=\frac{x+y+2z}{4}+\left|\frac{x-y}{4}\right|+\left|\frac{x+y-2z}{4}+\left|\frac{x-y}{4}\right|\right|$$
But this has nested absolute values, which I don't want.
I did find an approximate formula, which is exact when one of the variables is $0$ or when all are equal:
$$\max(|x|,|y|,|z|)\approx$$
$$-\tfrac13\Big(|x|+|y|+|z|\Big)\\+\tfrac16\Big(|x+y|+|x-y|+|x+z|+|x-z|+|y+z|+|y-z|\Big)\\+\tfrac16\Big(|x+y+z|+|x+y-z|+|x-y+z|+|x-y-z|\Big)$$
 A: The answer to Q1 is negative.
Let $x\in {\bf R}$ and $\alpha \in [0,1]$.
$$
max(x,\alpha x,0) = max(x,0) = {1\over 2} x + {1 \over 2} |x|.
$$
On the other hand, assuming Q1 true,
$$
max(x,\alpha x,0) = ({a_0+\alpha b_0})\ x + C(\alpha)\ |x|,
$$
where $C(\alpha)$ does not depend on $x$. Since the two functions $x\mapsto x$ and $x \mapsto |x|$ are linearly independent, we obtain
$$\forall \alpha \in [0,1], \ 
a_0+b_0\alpha = 1/2$$
and so $a_0 = 1/2$ and $b_0 = 0$.
Repeating the same argument with $max(\alpha x, x, 0)$, we end with $a_0 =0$, $b_0 = 1/2$, a contradiction.
A: For Q2:
Consider the set of points where it isn't differentiable. First, make sure there's no duplication in your sum by making sure (wlog) none of the terms are scalar multiples of each other. This guarantees the set where it isn't differentiable is just the union of the planes $a_i x + b_i y + c_i z = 0$. This would have to be equal to the set where $max(|x|,|y|,|z|)$ is not differentiable, which is equivalent to
$$(x,y,z) \in S \iff |x|=|y| \ge |z| \space \vee|x|=|z| \ge |y| \space \vee |y|=|z| \ge |x|$$
But this can't be made as a union of planes. Take a point where $|x| \ne |y| \ne |z|$. Clearly this isn't in $S$. But any plane through the origin will contain a point like this unless it is the plane $x=y$ or $x=z$ or $y=z$. And one can check that these planes aren't contained in $S$ either. E.g. $(0,0,1) \notin S$ so $x=y$ doesn't work.
So the sum would have to be empty. And clearly that doesn't work.
This also proves the answer to Q1, but @coudy already handled that.
