<s>[INST] 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.
 [/INST] Hint:
If all three numbers are negative, then:
$$ab > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm} ac > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm} bc > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm}$$
Therefore, if we multiply the three inequalities:
$$a^2b^2c^2 > \left(\frac{a+b+c}{3}\right)^6$$
Or equivalently:
$$\left(\sqrt[3]{abc}\right)^6 > \left(\frac{a+b+c}{3}\right)^6$$
Do you know any inequality you can use here do disprove this?
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