AUT_ABooklet_2020

Answer. Equality occurs if and only if $a=b=c=2$.

We set $x=a+1$, $y=b+1$ and $z=c+1$. Thus we have to show $$xyz\ge 27$$ subject to $$xyz=xy+yz+zx.$$ From the constraint we get $$xyz=xy+yz+zx\ge 3\sqrt[3]{x^2{y}^2{z}^2}$$ by using the inequality between the arithmetic and the geometric means of $xy$, $yz$ and $zx$. This is clearly equivalent to $xyz\ge 27$. Equality occurs if and only if $xy=yz=zx$, or, equivalently, $x=y=z$. By the constraint, this is equivalent to $x=y=z=3$ and finally $a=b=c=2$.