Ta có $-1< a;b;c< 1\Rightarrow \left\{\begin{matrix} \left ( 1+a \right )\left ( 1+b \right ) \left ( 1+c \right )> 0& \\ \left ( 1-a\right )\left ( 1-b \right )\left ( 1-c \right )> 0 & \end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix} abc+ab+bc+ca+a+b+c+1> 0 & \\ 1-abc+ab+bc+ca-a-b-c> 0 & \end{matrix}\right.$
$\Rightarrow 2+ab+bc+ca> 0 \Rightarrow 2+\left ( a+b+c \right )^{2}> a^{2}+b^{2}+c^{2}$
Mà a+b+c=0 nên $a^{2}+b^{2}+c^{2}< 2$