## Sunday, August 19, 2007

### When brain is too free..

`An interesting puzzle posted in Yahoo!Answers that I managed to solve, and gotten chosen as the best.===Let a, b, c be the lengths of the sides of a triangle. Show that:a/(b+c) + b/(a+c) + c/(a+b) < 2Now see if you can describe the shape of a triangle for which the above expression is very close to 2.=== Best Answer - Chosen by AskerStop here for a moment, as this is your chance to solve it before reading my answer below :)Somehow, I started with the last part first, when it is close to 2.So, using limits, try a triangle with one infinitely short side.lim_a->0 (a/(b+c) + b/(a+c) + c/(a+b))= lim_a->0 (0/(b+c) + b/c + c/b)but in such a triangle, where a->0, b->cit is actually lim_a->0_and_b->c (a/(b+c) + b/(a+c) + c/(a+b))= lim_a->0_and_b->c (0/(b+c) + b/c + c/b)= lim_a->0_and_b->c (0/(2c) + c/c + c/c)= 2Ok, now for the main question, to prove a/(b+c) + b/(a+c) + c/(a+b) < 2Let a < b < c,a/(b+c) < 1/2 because b+c > 2ac/(a+b) < 1 because in a triangle,             sum of lengths of two sides > the other side.What about, b/(a+c)? a < b, c > b but c < a+b, so c < 2b, so a+c < 3b, and thus b/(a+c) > 1/3 (Oh oh.. no use)Probably have to consider the first two terms together, we havea/(b+c) + b/(a+c) = (a^2+ac+b^2+bc)/(b+c)(a+c)= (a^2+ac+b^2+bc)/(ab+ac+bc+c^2)< 1because ac=ac, bc=bc, a^2 < ab, b^2 < c^2So, (a/(b+c) + b/(a+c) + c/(a+b)) < 2` 