A point inside a triangle

Question

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: $$b\ge EC \space\space\space\space and \space\space\space\space a\ge DC$$ $$b\cdot d_a\ge EC\cdot d_a \space\space\space\space and \space\space\space\space a\cdot d_b \ge DC \cdot d_b$$ $$b\cdot d_a+a\cdot d_b \ge EC\cdot d_a+DC \cdot d_b=DE \cdot R_C$$

Answer 1

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 $P'$.

Answer 2

$ad_a + bd_b + c_dc = 2\times Area$

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

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

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