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I am given a triangle $\triangle ABC$ with side lengths $a,b,c$ and a point $P$ inside it.
$R_A=PA$, $R_C=PC$, $R_C=PC$
the distances from point $P$ to the sides $BC, AC, AB$ are $d_a, d_b, d_c$ respectively.

How can I prove $b\cdot d_a+a\cdot d_b \leq c\cdot R_C$? I'd like to get a hint for where to start.

My attempt (I didn't find the solution but thats the closest I could get):

Firstly, I added points $D,E,F$ the projections of $P$ on sides $BC,AC,AB$ respectively. Also $BC=a,AC=b,AB=c$. Then: $$b\ge EC \space\space\space\space and \space\space\space\space a\ge DC$$ $$b\cdot d_a\ge EC\cdot d_a \space\space\space\space and \space\space\space\space a\cdot d_b \ge DC \cdot d_b$$ $$b\cdot d_a+a\cdot d_b \ge EC\cdot d_a+DC \cdot d_b=DE \cdot R_C$$

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  • $\begingroup$ I solved your problem. If you want to see my solution, show please your attempts. $\endgroup$ – Michael Rozenberg Jan 10 at 21:15
  • $\begingroup$ @MichaelRozenberg I edited what I found into the original post, although it's not what I need to prove. Is it a good direction? $\endgroup$ – aradarbel10 Jan 10 at 22:06
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The hint.

Make the following.

  1. Prove that: $ad_a+bd_b\leq cR_c;$

  2. Prove the previous inequality for any point $P$ inside the angle $ACB$;

  3. Take $P'$ symmetric to $P$ respect to bisector of $\angle ACB$ and write an inequality 1. for the point $P'$.

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  • $\begingroup$ I couldn't really get part 3, but you can't guarantee that $P'$ is going to be inside the triangle... $\endgroup$ – aradarbel10 Jan 11 at 10:32
  • $\begingroup$ @aradarbel10 Yes, of course, but $P'$ is placed inside the angle ACB and we can use 2. $\endgroup$ – Michael Rozenberg Jan 11 at 11:05
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$ad_a + bd_b + c_dc = 2\times Area$

$d_c + R_c$ is greater that the shortest line from from $C$ to side $c$

$(d_c + R_c)c\ge 2\times Area$

$(d_c + R_c)c \ge ad_a + bd_b + c_dc\\ cR_c \ge ad_a + bd_b$

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  • $\begingroup$ You solved 1. in my hint. It's an easiest part of the proof. See please my post. $\endgroup$ – Michael Rozenberg Jan 11 at 6:41

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