Estonian Mathematical Olympiad

The given equations are equivalent to $2a=b^2=\frac{c}{2}$. So $b^2$ is divisible by $2$, therefore $b$ is also divisible by $2$. Since the product $abc$ is divisible by $3$, one of the numbers $a$, $b$ and $c$ must be divisible by $3$. If $a$ is divisible by $3$, then $b^2$ is also divisible by $3$, therefore $b$ must be divisible by $3$. If $c$ is divisible by $3$, then $b^2$ must also be divisible by $3$. So in all cases $b$ is divisible by $3$. Therefore $b$ is divisible by both $2$ and $3$, so $b$ must be at least $2\cdot 3=6$. If $b=6$, then $a=\frac{b^2}{2}=18$ and $c=2b^2=72$ and $a+b+c=96$. If $b$ is larger, then $b^2$ and also $a$ and $c$ would be larger, hence the sum found is the smallest possible.