Hope that I didn't do anything stupid. At first, suppose the circle $ R_{A} $ tangents at the point $ D $ to the side $ BC $ By using spiral similarity, we know that $ AD $ is the angle bisector. As the result, we get $ 2R_{A} \ge l_{A} $. Similarly, $ 2R_{B} \ge l_{B},2R_{C} \ge l_{C} $ Finally, notice that we have a well-known lemma, that is, \[ \frac{1}{l_{A}}+\frac{1}{l_{B}}+\frac{1}{l_{C}} \le \frac{1}{r} \]

A Funny Problem Notice that we have $ 2R_{A} \ge h_{A}, 2R_{B} \ge h_{B}, 2R_{C} \ge h_{C} $ And the identity $ \frac{1}{h_{A}}+\frac{1}{h_{B}}+\frac{1}{h_{C}} = \frac{1}{r} $, so we're done!

BSJL wrote: A Funny Problem Notice that we have $ 2R_{A} \ge h_{A}, 2R_{B} \ge h_{B}, 2R_{C} \ge h_{C} $ And the identity $ \frac{1}{h_{A}}+\frac{1}{h_{B}}+\frac{1}{h_{C}} = \frac{1}{r} $, so we're done! Sorry but can you explain me why these inequalities hold?

Let $\Gamma$ be the circumcircle of $\triangle ABC$, and extend the altitude from $A$ to intersect $\Gamma$ at $X$. Now just note that $AX$ is a chord and thus has length at most that of the diameter of the circumcircle, which is $2R$. Similar reasoning can be used for $h_B$ and $h_C$.

Let$Oa$ be center of circle with radius $Ra$ and $Pa$ point where $Ra$ touches $BC$ and let $AA'$be altitude from$A$ if $Oa'$ is porjection of $Oa$ to $AA'$ we have $AOa \ge AOa'$ and $OaPa=Oa'A'$ which now implies $Ra \ge ha/2$ so now $1/Ra \le 2/ha$ now by mulitplying the indentity by $S$ the surface of trinagle we have $\sum 1/Ra\le \sum 2S/ha=a+b+c=P*2/r$ . Done.

http://www.cut-the-knot.org/triangle/DorinAndrica.shtml

Let $I$ be the incenter of $\triangle ABC$ Let $A'$ the intersection of $AI$ and $BC$ and $B'$,$C'$ defined similary It is easy to see that $2R_{A} \ge AA' \ge h_{A}$ similary $2R_{B} \ge BB' \ge h_{B}$ and $2R_{C} \ge CC' \ge h_{C}$......$(1)$ Then from the $(1)$ ... $\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C}\leq\frac{2}{h_{A}}+\frac{2}{h_{B}}+\frac{2}{h_{C}}$. Then we apply the following equation: $ \frac{r}{h_A} = \frac{a}{a+b+c} $ we get it $ \frac{1}{h_{A}}+\frac{1}{h_{B}}+\frac{1}{h_{C}} = \frac{1}{r} $ Then finally: $\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C}\leq\frac{2}{h_{A}}+\frac{2}{h_{B}}+\frac{2}{h_{C}} = \frac{2}{r} $ l.q.q.d

Here my solution: It’s equivalent to $ cos(\frac{B-C}{2})^2sinA+ cos(\frac{C-A}{2})^2sinB+ cos(\frac{A-B}{2})^2sinC \le 4cos(\frac{A}{2}) cos(\frac{B}{2}) cos(\frac{C}{2}) $.Which is true ,because we know that $ sinA+sinB+sinC=4 cos(\frac{A}{2}) cos(\frac{B}{2}) cos(\frac{C}{2}) \ge cos(\frac{B-C}{2})^2sinA+ cos(\frac{C-A}{2})^2sinB+ cos(\frac{A-B}{2})^2sinC $.

BSJL wrote: Hope that I didn't do anything stupid. At first, suppose the circle $ R_{A} $ tangents at the point $ D $ to the side $ BC $ By using spiral similarity, we know that $ AD $ is the angle bisector. As the result, we get $ 2R_{A} \ge l_{A} $. Similarly, $ 2R_{B} \ge l_{B},2R_{C} \ge l_{C} $ Finally, notice that we have a well-known lemma, that is, \[ \frac{1}{l_{A}}+\frac{1}{l_{B}}+\frac{1}{l_{C}} \le \frac{1}{r} \]