An interesting puzzle for you... (corrected version)?

Question

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.

Any answers, and any explanations? Sorry about the wrong version posted here - I'm glad someone pointed out the notation was wrong. This is the correct version of the puzzle, which should make more sense!

Have fun with this one.

Answer 1

Multiplying the three permutations of the triangle inequality:

0 < a + b - c , etc.

together gives:

0 < (a+b-c)(a+c-b)(b+c-a)

-->

0 < bc^2 + ac^2 + cb^2 + ab^2 + ca^2 + ba^2
-2abc -c^3 -b^3 -a^3

-->

3abc < bc^2 + ac^2 + cb^2 + ab^2 + ca^2 + ba^2
+abc -c^3 -b^3 -a^3

-->

3abc < (b+c)(a+c)(a+b)[2 - a/(b+c)-b/(a+c)-c/(a+b)]

-->

a/(b+c) + b/(a+c) + c/(a+b) < 2 - 3abc/((b+c)(a+c)(a+b))

So the inequality holds. It approaches an equation only if a,b, or c --> 0 as mentioned in a previous answer.

Answer 2

Pick one side to be almost 0, and 2 sides to be almost infinity.

One of the above expressions will be almost 0 and the other 2 will be almost 1.

The total is of course almost 2

Answer 3

Though I cannot prove the above, by symmetry the above is maximum when b=a = c and the result is 1.5
2 is very far

Answer 4

when the triangle is equilateral, it gets to 1.5.... far than 2.

Answer 5

How did you find this? I dont know the answer yet, but I wil find out!!!!