THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

First solution. If $a+b+c\le 0$, the inequality is obviously satisfied, with equality occurring if and only if $a=b=c=0$. If $a+b+c>0$, then $abc<0$. It is not possible for all the variables to be negative (their sum would be negative), therefore one of them is negative and the other two are positive. We may assume that $a<0<b,c$.

Put $x=-a$. Then $-x+b+c=xbc>0$ and $(a^2+b^2+c^2{)}^2=(x^2+b^2+c^2{)}^2\ge (xb+xc+bc{)}^2\ge 3(xb\cdot xc+xc\cdot bc+bc\cdot xb)=3xbc(x+b+c)=3(-x+b+c)(x+b+c)\ge 9(-x+b+c{)}^2$. As $-x+b+c>0$, the last inequality comes to $x+b+c\ge 3(-x+b+c)$, i.e. to $x\ge \frac{b+c}{2}$. But $x=\frac{b+c}{1+bc}\ge \frac{b+c}{2}$ because $bc\le 1$. In conclusion, we obtain $a^2+b^2+c^2\ge 3(-x+b+c)=3(a+b+c)$, with equality (in the case $a<0<b,c$) if and only if $x=b=c$ and $bc=1$, i.e., if $a=-1$, $b=c=1$. To conclude, we have equality if $(a,b,c)\in \{(0,0,0),(-1,1,1),(1,-1,1),(1,1,-1)\}$.

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Alternative solution.

Second solution. If $a+b+c\le 0$, the inequality is obviously satisfied, with equality occurring if and only if $a=b=c=0$. If $a+b+c>0$, then $abc<0$. From the AM-GM inequality, $a^2+b^2+c^2\ge 3\sqrt[3]{(abc{)}^2}\ge 3|abc|=-3abc=3(a+b+c)$, with equality if $a^2=b^2=c^2$, $abc\le 0$ and $|abc|=1$, which leads to $(a,b,c)\in \{(0,0,0),(-1,1,1),(1,-1,1),(1,1,-1)\}$.