1. The obtuse-angled triangle $ABC$ has sides of length $a,b$ and $c$ opposite the angles $\angle A, \angle B$ and $\angle C$ respectively. Prove that $$a^3 \cos A + b^3 \cos B + c^3 \cos C \lt abc.$$

No real idea what to do, so any contributions are appreciated.

  • $\begingroup$ Please use this guide for formatting mathematics on this site. It makes things much easier for everyone who reads your question. For instance, I assume that when you write "a3 cos A", you really mean $a^3\cos A$, but for all I know it might be $a \cdot 3\cos A$, or even $a_3 \cos A$, and there is no way for me to tell. $\endgroup$ – Arthur Apr 6 '15 at 19:41
  • $\begingroup$ I saw this page. $\endgroup$ – mathlove Apr 6 '15 at 19:43
  • $\begingroup$ Ok thanks. Any ideas on the question? $\endgroup$ – MadChickenMan Apr 6 '15 at 19:49
  • $\begingroup$ use the theorem of cosine $\endgroup$ – Dr. Sonnhard Graubner Apr 6 '15 at 20:15
  • $\begingroup$ I've tried, but not really got anywhere. $\endgroup$ – MadChickenMan Apr 6 '15 at 21:41

it is equivalent to $\frac{\left(a^2-b^2-c^2\right) \left(a^2+b^2-c^2\right) \left(a^2-b^2+c^2\right)}{2 a b c}>0$

  • $\begingroup$ I'm afraid I don't really see how you can't to this conclusion. $\endgroup$ – MadChickenMan Apr 7 '15 at 10:29
  • $\begingroup$ Sorry, meant to say came* $\endgroup$ – MadChickenMan Apr 7 '15 at 11:08
  • $\begingroup$ Ah I think I may know what you mean, but tell me: how do you make these factorisations. There are so many terms, is it simply a case of having seen the factorisation before? $\endgroup$ – MadChickenMan Apr 7 '15 at 11:29
  • $\begingroup$ i have solved many of such problems in our math-circle in Leipzig with my students two of them participated at the last IMO in Southafrica with a silver medall $\endgroup$ – Dr. Sonnhard Graubner Apr 7 '15 at 13:08
  • $\begingroup$ As someone who is familiar with IMO questions then, may I ask how this question and other BMO questions (if you are familiar with those) compare. $\endgroup$ – MadChickenMan Apr 7 '15 at 15:39

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