Find all triplets of real numbers $a, b, c$ such that $$|ax+by+cz|+|bx+cy+az|+|cx+ay+bz|=|x|+|y|+|z|$$ is true for all real number $x,y,z$.
I tried a lot but couldn't solve the problem.
$x=y=z=1 \implies |a+b+c|=1 $
and $x=y=0,z=1\implies |a|+|b|+|c|=1$
But I have no idea how to use this to find all possible values of $a, b $ and $c$. How can I proceed further?
You have made great progress as quoted above.
Towards a contradiction, suppose at least two of $a,b,c$ are nonzero. WLOG, suppose both $a$ and $b$ are nonzero. If $x=a$, $y=-b$, $z=0$, then
$\quad |ax+by+cz| = |a^2-b^2| < a^2 + b^2$
$\quad |bx+cy+az| = |ba-cb| \le |b||a| + |c||b|$
$\quad |cx+ay+bz| = |ca-ab| \le |c||a| + |a||b|$
Adding the inequalities, we get $$\begin{aligned} &\quad |ax+by+cz|+|bx+cy+az|+|cx+ay+bz|\\ &< (|a|+|b|)(|a|+|b|+|c|)\\ &= |a| + |b| \\ &= |x|+|y|+|z|,\\ \end{aligned}$$ which contradicts with the given condition. Hence, at most one of $a,b,c$ is nonzero.
Since $|a|+|b|+|c|=1$, we know $(a,b,c)$ must be one of the following six triplets. $(\pm1, 0, 0), (0, \pm1, 0), (0, 0, \pm1).$
It is straightforward to verify each of the six triplets satisfies the given condition.
Exercise 1. Find all pairs of real numbers $a, b$ such that $|ax+by|+|bx+ay|=|x|+|y|$ for all real number $x$ and $y$.
Exercise 2. Find all quadruplets of real numbers $a, b, c,d$ such that $|aw+bx+cy+dz|$ $+|bw+cx+dy+az|$ $+|cw + dx+ay+bz|$ $+|dw +ax+by+cz|$ $=|w|+|x|+|y|+|z|$ is true for all real number $w,x,y,z$