3

This is my proof. I'm not sure if it is true but perhaps someone would tell me yes or (no and why).

Since $(a^2+b^2), (b^2+c^2), (c^2+a^2)$ are greater than or equal to zero, then $(a^2+b^2)(b^2+c^2)(c^2+a^2) \geq 0$. I'll spare you the distribution and say this, $(a^2+b^2)(b^2+c^2)(c^2+a^2) + 8a^2b^2c^2 \geq 8a^2b^2c^2$

Done.

Aryabhata
  • 82,206
JetRex
  • 337
  • 5
    No, that does noy work. A hint: $x^2+y^2\ge 2xy$. – Andrés E. Caicedo Dec 17 '13 at 02:20
  • You actually need to show $(a^2+b^2)(b^2+c^2)(c^2+a^2)\geq 8a^2b^2c^2$... I do not understand why would $(a^2+b^2)(b^2+c^2)(c^2+a^2) + 8a^2b^2c^2 \geq 8a^2b^2c^2$ imply $(a^2+b^2)(b^2+c^2)(c^2+a^2)\geq 8a^2b^2c^2$ –  Dec 17 '13 at 02:21
  • That solution would work for any number greater than 0. Suppose we wanted to prove 5>6. Then 5>0 and therefore 5+6>6. But this doesn't prove 5>6. – Asinomás Dec 17 '13 at 02:23
  • +1 for @AndresCaicedo hint.... you need to apply that thrice.... Good luck! –  Dec 17 '13 at 02:24

1 Answers1

4

rewriting what andres said : $(a^2+b^2)\geq2ab,(b^2+c^2)\geq2bc,(c^2+a^2)\geq2ca$.

therefore $(a^2+b^2)(b^2+c^2)(c^2+a^2) \geq (2ab)(2bc)(2ca) = 8a^2 b^2 c^2$

Asinomás
  • 105,651
  • I have down voted this... It would not be a better idea to write complete solution if some other person has given hints just before ten minutes... you could have given some time... –  Dec 17 '13 at 02:33
  • ok, you may be right, but I am completely against the romantic idea so many people here have about education, seeing a solution isn't that bad, besides if people are asking then they shouldn't get angry when they have their questions answered. Also, the interesting part of the problem had already been given up, all I did was write it up. if someone is not able to answer the problem with that hint they should go back a little anyways. – Asinomás Dec 17 '13 at 02:36
  • seeing a solution is not that bad... the person who asked the question should see the solution by trying a bit hard with given hint... any way it would be their personal choice... –  Dec 17 '13 at 02:41
  • I have revisited this and see the solution is elementary. I suppose with a bit of hard work I would have arrived to a solution. At the time, however, I had a bit of a penchant for working on homework at the last minute. – JetRex Jun 10 '16 at 16:40