Croatian Junior Mathematical Olympiad

Let us denote $x=a-1$, $y=b-1$ and $z=c-1$. Then $x,y,z\ge 0$, and the given inequality easily transforms into $$3xyz+xy+yz+zx\ge 0,$$ which is true since all addends on the left-hand side are non-negative.

The equality is obtained if and only if $xyz=xy=yz=zx=0$, which is true if and only if at least two numbers among $x,y$ and $z$ are equal to $0$, i.e. if and only if at least two numbers among $a,b$ and $c$ are equal to $1$.