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.

  • $\begingroup$ Have you considered adding them up and seeing if they are a square ? $\endgroup$ – Belgi Aug 16 '14 at 17:42


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?

  • $\begingroup$ @Sudhanshu Look for the Arithmetic-Geometric Median Inequality en.wikipedia.org/wiki/… It says that: $$\frac{a+b+c}{3} \geq \sqrt[3]{abc}$$ $\endgroup$ – Darth Geek Aug 16 '14 at 17:50
  • $\begingroup$ Oh, how can I forget that...Thanks btw. $\endgroup$ – Sudhanshu Aug 16 '14 at 17:54
  • $\begingroup$ But since we assume these numbers to be negative how can we apply A.M > G.M ? $\endgroup$ – Sudhanshu Aug 16 '14 at 18:04
  • $\begingroup$ If we assume those three numbers are negative, then they contradict the "AM>GM" inequality wich we know to be true. Therefore our assumption was incorrect, i.e. at least one of those numbers was not negative. This sort of proof is called "proof by contradiction" or "reductio ad absurdum". You start with a hypothesis and you end up in a contradiction, wich means that the hypothesis was false. $\endgroup$ – Darth Geek Aug 16 '14 at 18:14
  • $\begingroup$ But I think the A.M > G.M rule is applicable only for positive numbers $\endgroup$ – Sudhanshu Aug 16 '14 at 18:15

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

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

  • $\begingroup$ @Sudhanshu, How about this? $\endgroup$ – lab bhattacharjee Aug 16 '14 at 18:24
  • $\begingroup$ Could you explain how you took 3 as common out of that Sum sign? $\endgroup$ – Sudhanshu Aug 16 '14 at 18:34
  • $\begingroup$ You can also observe that $\sum[a^2+b^2+c^2-ab-bc-ca]\ge0$ is just Cauchy -Schwarz. $\endgroup$ – N. S. Aug 16 '14 at 18:38
  • $\begingroup$ @Macavity, Thanks for your observation $\endgroup$ – lab bhattacharjee Aug 17 '14 at 6:02

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