Geometric inequality $\frac{R_a}{2a+b}+\frac{R_b}{2b+c}+\frac{R_c}{2c+a}\geq\frac{1}{\sqrt3}$

Let $P$ be a point inside $\triangle{ABC}$.

Let $AP=R_a$, $BP=R_b$, $CP=R_c$, $AB=c$, $BC=a$ and $CA=b$. Prove that:

$$\frac{R_a}{2a+b}+\frac{R_b}{2b+c}+\frac{R_c}{2c+a}\geq\frac{1}{\sqrt3}$$

I tried to use C-S, but without success because the equality case is $a=b=c$

and $P$ is a center of a triangle, but by the way, if $P\equiv C$ we obtain: $$\sum\limits_{cyc}\frac{R_a}{2a+b}=\frac{b}{2a+b}+\frac{a}{2b+c}>\frac{b}{2a+b}+\frac{a}{a+3b}>\frac{1}{\sqrt3},$$

where the last inequality is true and strong enough. It says that it's very difficult to use C-S here. Thank you!

Here is an idea that I think could work.

$\textbf{Disclaimer: }$This is not a solution, but rather a transformation of the problem into a more 'manageable' form.

We locate the triangle in the complex plane with vertices $A= x_1+\Bbb i x_2,\; B= y_1+\Bbb i y_2$ and $C= z_1+\Bbb i z_2.$ Whitout loss of generalization $P=0.$ We now have

$$a=|B-C|,\; b=|C-A|,\;c=|A-B|,\; R_a= |A|,\; R_b= |B|,\; R_c= |C|.$$ In order to guarantee that $P$ lies inside $\triangle ABC,$ it is necessary and sufficient the existence of scalars $\alpha,\beta,\gamma$ such that:

$$\alpha,\beta,\gamma\geq 0,\;\alpha+\beta+\gamma=1, \; \alpha A+ \beta B+\gamma C=0.$$ The problem is now equivalent to

$$\min f(A,B,C,\alpha,\beta,\gamma)= \frac{|A|}{2|B-C| + |C-A|} + \frac{|B|}{2|C-A| + |A-B|}+ \frac{|C|}{2|A-B| + |B-C|},$$

$$s.t \;\; \alpha,\beta,\gamma\geq 0,\;\alpha+\beta+\gamma=1, \; \alpha A+ \beta B+\gamma C=0.$$ Furthermore, since $f$ is homogeneous, we can assume whitout loss of generality that $|A|^2+|B|^2+|C|^2=1,$(or $|A|+|B|+|C|=1$) so add this constraint to the optimization problem. Now the problem is stated in real terms rather than in geometrical terms, so should make more sense to apply C-S or other known inequalities. Best of lucks!!