if one of $a,b,c$ is $0$ then $a+b+c=0$ if $abc \ge 0 $ which means $a^2+b^2+c^2 \ge 0 \ge 3(a+b+c) $ then $abc<0$ if $a,b,c$ are all negative then the sum is negative then we have $a,b>0$ let $c'=-c>0$ now $1\ge abc'>0$ or $1\ge abc'^{\frac{1}{3}}>0$ note that $a^2+b^2+c'^2 \ge 3(a^2b^2c'^2)^\frac{1}{3} \ge 3abc'^{\frac{1}{3}}*(a^2b^2c'^2)^\frac{1}{3} =3abc'=3(a+b-c') $ equality when $ (a,b,c)=(1,1,-1)$