1
$\begingroup$
  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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.