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) < 2

Now see if you can describe the shape of a triangle
for which the above expression is very close to 2.

=== Best Answer - Chosen by Asker

Stop 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->c
it 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)
= 2

Ok, now for the main question,
to prove a/(b+c) + b/(a+c) + c/(a+b) < 2
Let a < b < c,
a/(b+c) < 1/2 because b+c > 2a
c/(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 have
a/(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)
< 1
because ac=ac, bc=bc, a^2 < ab, b^2 < c^2

So, (a/(b+c) + b/(a+c) + c/(a+b)) < 2

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