Proving that a number is non-negative?

Question

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.

Answer 1

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?

Answer 2

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

If each $(a+b+c)^2-3bc<0,$ $$\sum [(a+b+c)^2-3bc]<0$$