# BMO1 2008/09 Question 6 Trigonometry Problem

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.

• 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. – Arthur Apr 6 '15 at 19:41
• I saw this page. – mathlove Apr 6 '15 at 19:43
• Ok thanks. Any ideas on the question? – MadChickenMan Apr 6 '15 at 19:49
• use the theorem of cosine – Dr. Sonnhard Graubner Apr 6 '15 at 20:15
• I've tried, but not really got anywhere. – MadChickenMan Apr 6 '15 at 21:41

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