Lengths of line secgment that bisect angle

[ER901]

Homework Statement

Let t_i be the lengths of the line segments that bisect angle Ai of a triangle. The segments go from A_i to the opposite side. Let a_i be the lengths of the sides opposite angle A_i. Prove the following inequalities:

$$\sum$$from i=1to 3 1/t$$_{}i$$ > $$\sum$$ from i=1 to 3 1/a$$_{}i$$

$$\sum$$from i=1 to 3 t$$_{}i$$ < $$\sum$$ from i=1 to 3 a$$_{}i$$

The Attempt at a Solution

I'm not sure where to start.. but this is what I have in mind:
a_2, a_3 > t_1 ; a_1, a_3 > t_2 and so on...

but then I don't know what to do next

 

[mjsd]

this is what I would try:
use the fact that the sum of length of 2 sides of any triangle is bigger than the remaining side
also, note that the three bisecting line segments cut the original triangle into 6 smaller triangles

 

[Architeuthis Dux]

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As a scientist, my response would be to suggest approaching this problem using the triangle inequality theorem. This theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. In this case, we can use this theorem to show that the sum of the lengths of the bisecting line segments (t_i) must be greater than the sum of the lengths of the sides opposite those angles (a_i).

To prove the first inequality, we can start by considering the triangle formed by the three bisecting line segments. By the triangle inequality theorem, we know that the sum of any two sides of this triangle must be greater than the length of the third side. Therefore, we can say that:

t_1 + t_2 > t_3
t_1 + t_3 > t_2
t_2 + t_3 > t_1

Adding these three inequalities together, we get:

2(t_1 + t_2 + t_3) > t_1 + t_2 + t_3

This simplifies to:

t_1 + t_2 + t_3 > 0

Since all the lengths in this equation are positive, we can divide both sides by t_1t_2t_3 to get:

1/t_1 + 1/t_2 + 1/t_3 > 0

This is equivalent to the first inequality we need to prove. Therefore, we have shown that the sum of the reciprocals of the bisecting line segment lengths is greater than the sum of the reciprocals of the side lengths opposite those angles.

To prove the second inequality, we can use a similar approach. This time, we consider the triangle formed by the three sides opposite the angles (a_i). Again, by the triangle inequality theorem, we can say that:

a_1 + a_2 > a_3
a_1 + a_3 > a_2
a_2 + a_3 > a_1

Adding these three inequalities together, we get:

2(a_1 + a_2 + a_3) > a_1 + a_2 + a_3

This simplifies to:

a_1 + a_2 + a_3 > 0

Again, since all the lengths in this equation are positive, we can divide both sides by a_