Proving that a number is non-negative? The numbers $a$,$b$ and $c$ are real.
 Prove that at least one of the three numbers 
$$(a+b+c)^2 -9bc \hspace{1cm} (a+b+c)^2 -9ca \hspace{1cm} (a+b+c)^2-9ab$$
is non-negative.
Any hints would be appreciated too.

$$\sum [(a+b+c)^2-9bc]=3\sum[a^2+b^2+c^2-ab-bc-ca]=\frac32\sum (a-b)^2\ge0$$
If each $(a+b+c)^2-3bc<0,$  $$\sum [(a+b+c)^2-3bc]<0$$
