I have been struggling with this problem:
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.
Let $l_a$ be the length of the angle bisector of the angle $\hat A$ in the given triangle. Using usual notations, $$ l_a^2 =\frac 1{(b+c)^2}\; 4bc\; p(p-a)\ . $$ (Here, $a,b,c$ are the sides of $\Delta ABC$, and $p=\frac 12(a+b+c)$ is the half-perimeter.) Then we have $$ \tag{$*$} \frac 2{l_a}=\frac {b+c}{\sqrt{bc\; p(p-a)}} \color{red}{\ge} \frac{b+c}{bc}=\frac 1b+\frac 1c\ . $$ Indeed, the red inequality is successively equivalent to... $$ \begin{aligned} \sqrt{bc\; p(p-a)} &\le bc\ ,\\ p(p-a) &\le bc\ ,\\ ((b+c)+a)((b+c)-a) &\le 4bc\ ,\\ (b+c)^2-a^2 &\le 4bc\ ,\\ (b+c)^2-4bc -a^2&\le 0\ ,\\ (b-c)^2 -a^2&\le 0\ ,\\ (b-c-a)(b-c+a)&\le 0\ . \end{aligned} $$ And the last inequality is true. Now use cyclic cousins of $(*)$, and add all three inequalities.