Prove that the sum of the reciprocals of the lengths of the interior angle bisectors of a triangle is greater than the sum of the reciprocals of the lengths of the sides of the triangle

beefypy

beefypy

Answered question

2022-10-14

Prove that the sum of the reciprocals of the lengths of the interior angle bisectors of a triangle is greater than the sum of the reciprocals of the lengths of the sides of the triangle

I have tried different approaches to solve it here are some of them :

-The relation between the measures of angles in a triangle and the lengths of sides

-Trying triangle inequality on different triangles

-Using some formulas and inequalities involving angle bisectors and sides

Unfortunately, I wasn't able to solve it.

I don't want the full solution I just need some hints and suggestions on how to solve this problem.

Answer & Explanation

Amadek6

Amadek6

Beginner2022-10-15Added 21 answers

Let l a be the length of the angle bisector of the angle A ^ in the given triangle. Using usual notations,
l a 2 = 1 ( b + c ) 2 4 b c p ( p a )   .
(Here, a,b,c are the sides of Δ A B C, and p = 1 2 ( a + b + c ) is the half-perimeter.) Then we have
( ) 2 l a = b + c b c p ( p a ) b + c b c = 1 b + 1 c   .
Indeed, the black inequality is successively equivalent to...
b c p ( p a ) b c   , p ( p a ) b c   , ( ( b + c ) + a ) ( ( b + c ) a ) 4 b c   , ( b + c ) 2 a 2 4 b c   , ( b + c ) 2 4 b c a 2 0   , ( b c ) 2 a 2 0   , ( b c a ) ( b c + a ) 0   .
And the last inequality is true. Now use cyclic cousins of ( ), and add all three inequalities.

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