Good problem ! Thank you !

Use AM-GM for three numbers , we have :

a² + b² + c² \(\ge3.\sqrt[3]{a^2.b^2.c^2}\)

\(\Leftrightarrow\left(\dfrac{a^2+b^2+c^2}{3}\right)^3\ge a^2b^2c^2\)

\(\Leftrightarrow\left(\dfrac{3}{3}\right)^3\ge a^2b^2c^2\)

\(\Leftrightarrow1\ge a^2b^2c^2\)

\(\Leftrightarrow1\ge abc\)

\(\Leftrightarrow-abc\le1\)              (1) 

Now , we have to prove this .

(|a| + |b| + |c|)² \(\le\) 3(a² + b² + c²)

\(\Leftrightarrow\left|a\right|^2+\left|b\right|^2+\left|c\right|^2+2.\left|a\right|.\left|b\right|+2.\left|b.\right|.\left|c\right|+2.\left|c\right|.\left|a\right|\le3a^2+3b^2+3c^2\)

\(\Leftrightarrow a^2+b^2+c^2+2.ab+2.bc+2.ca\le3a^2+3b^2+3c^2\)

\(\Leftrightarrow2.ab+2.bc+2.ca\le2a^2+2b^2+2c^2\)

\(\Leftrightarrow0\le2a^2+2b^2+2c^2-2ab-2bc-2ca\)

\(\Leftrightarrow0\le\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)           (right)

So , \(\left(\left|a\right|+\left|b\right|+\left|c\right|\right)^2\le3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow\left(\left|a\right|+\left|b\right|+\left|c\right|\right)^2\le3.3=9\)

But |a| + |b| + |c| \(\ge\) 0

\(\Rightarrow\left|a\right|+\left|b\right|+\left|c\right|\le3\)                        (2)

From (1) and (2) 

=> |a| + |b| + |c| - abc \(\le\) 4

Equality occurs when and only when in three numbers a,b,c

+ There are 2 numbers equals 1

+ 1 number equals -1