Refinement of $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$

Question

It's a refinement of Prove $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$ .

It's a attempt to find a simple proof of this fact . Now the refinement :

Let $1\leq a\leq 2$ and $x\in[0,3-a]$ then define :

$$f(x,a)= \left(\frac{\frac{(1+a)}{(x+a)}+\frac{(x+1)}{(3-x-a+x)}+\frac{(3-x-a+1)}{(3-x+a-a)}}{3}\right)^{\frac{2}{5}}$$

And :

$$g(x,a)= \left(\frac{\frac{(x+a)}{(1+a)}+\frac{(3-x-a+x)}{(1+x)}+\frac{(3-x-a+a)}{(3-x+1-a)}}{3}\right)^{-\frac{2}{5}}$$

then we have :

$$\left(\frac{(1+a)}{(x+a)} \right)^{\frac25}+\left(\frac{(x+1)}{(3-x-a+x)} \right)^{\frac25}+\left(\frac{(3-x-a+1)}{(3-x+a-a)}\right)^{\frac25} \geqslant\frac{195}{100}f(x,a)+\frac{105}{100}g(x,a) \geqslant 3$$

My refinement is based on two observation :

With Jensen's inequality we have $x,y,z>0$ :

$$3\left(\frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{3}\right)^{-\frac{2}{5}}\leq x^{\frac{2}{5}}+y^{\frac{2}{5}}+z^{\frac{2}{5}}\leq 3\left(\frac{x+y+z}{3}\right)^{\frac{2}{5}}$$

For the rest I play with the coefficients to give the refinement.

For the RHS we put :

$$u=\frac{(1+a)}{(x+a)}$$

We make the same things for the others fractions and I think we can prove a more general inequality for $u,v,w$ . For the LHS I have tried to put on the same denominator but currently I don't see any good issue .

Update 13/12/2020:

We want to show a more general inequality with some constraints on it ,so we have the inequality :

$$x^{\frac{2}{5}}+y^{\frac{2}{5}}+z^{\frac{2}{5}}\geq \frac{105}{100}\left(\frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{3}\right)^{-\frac{2}{5}}+\frac{195}{100}\left(\frac{x+y+z}{3}\right)^{\frac{2}{5}}\quad(1)$$

Due to homogeneity we can put $u=\frac{x}{z}$ and $v=\frac{y}{z}$ and get :

$$1+u^{\frac{2}{5}}+v^{\frac{2}{5}}\geq \frac{105}{100}\left(\frac{\frac{1}{1}+\frac{1}{u}+\frac{1}{v}}{3}\right)^{-\frac{2}{5}}+\frac{195}{100}\left(\frac{1+u+v}{3}\right)^{\frac{2}{5}}\quad(I)$$

Now we make the subsitution $u=p^3$ and $v=q^3$ we get :

$$1+p^{\frac{6}{5}}+q^{\frac{6}{5}}\geq \frac{105}{100}\left(\frac{\frac{1}{1}+\frac{1}{p^3}+\frac{1}{q^3}}{3}\right)^{-\frac{2}{5}}+\frac{195}{100}\left(\frac{1+p^3+q^3}{3}\right)^{\frac{2}{5}}$$

Now we use Jensen's inequality we have :

$$p^{\frac{6}{5}}+q^{\frac{6}{5}}\geq 2\left(\frac{p+q}{2}\right)^{\frac{6}{5}}$$

Remains to show the new bound : $$1+2\left(\frac{p+q}{2}\right)^{\frac{6}{5}}\geq \frac{105}{100}\left(\frac{\frac{1}{1}+\frac{1}{p^3}+\frac{1}{q^3}}{3}\right)^{-\frac{2}{5}}+\frac{195}{100}\left(\frac{1+p^3+q^3}{3}\right)^{\frac{2}{5}}$$

And now I'm stuck ...My last idea was to use the Gauss identity wich states for reals $a,b,c$ :

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

Without success ! So I cannot find currently constraints on $x,y,z$ to have inequality $(1)$

For information we have the 2 variables version of the inequality ($x>0$):

$$\frac{2}{3}\left(\left(\frac{195}{100}\right) \left(\frac{1+x}{2}\right)^{\frac{2}{5}}+\frac{105}{100} \left(\frac{1+\frac{1}{x}}{2}\right)^{\frac{-2}{5}}\right)\leq 1+x^{\frac{2}{5}}$$

update : 14/12/2020

Starting from $(I)$ we can use Ji Chen's lemma to give some constraints . To works we need to split the coefficient $\frac{195}{100}$ into $1+\frac{95}{100}$ to have three variables . Finally I have tried for the LHS $uvw's$ method to find some others constraints .

Question :

How to show the refinement ?Is it also true for $2\leq a<3$?

Thanks !