Prove $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$ $a,b,c >0$ and $a+b+c=3$, prove
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$$
I try to apply AM-GM
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3\cdot \sqrt[3]{\left(\frac{a+1}{a+b} \right)^{\frac25}\left(\frac{b+1}{b+c} \right)^{\frac25}\left(\frac{c+1}{c+a} \right)^{\frac25}}$$
Thus it remains to prove
$$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1   $$ with the condition $a+b+c=3.$
 But I found the counter example for $$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1   $$ :(
 A: Only a partial answer.
Assuming $a\le b\le c$, then we have $0<a\le1$ and $1\le c<3$. Now we have two cases, $b\le1$ and $b>1$. The case $b\le1$ is easy to deal with.

Assuming $0<a\le b\le1\le c<3$, we have
\begin{align}
\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \ge 1
&\iff (a+1)(b+1)(4-a-b)\ge(a+b)(3-a)(3-b)\\
&\iff {\left(2-a-b\right)} {\left(1-a\right)} {\left(1-b\right)}\ge0.
\end{align}

Now assuming $0<a\le1\le b \le c<3$. Below is not an answer but only an analysis.
We want to know why this case is difficult, and why $\frac25$ is important.
Using the series expansion of $\exp$, and let $A = \operatorname{diag}\left(\ln\left(\frac{a+1}{a+b} \right),\ln\left(\frac{b+1}{b+c}\right),\ln\left(\frac{c+1}{c+a} \right)\right)$, then we have
\begin{align}
LHS 
&= \sum_{n=0}^\infty\left(\frac25\right)^n\frac{\operatorname{tr}A^n}{n!}\\
&= 3 + \frac25\operatorname{tr}A + \frac12\cdot \left(\frac25\right)^2\operatorname{tr}A^2+ \frac16\cdot \left(\frac25\right)^3\operatorname{tr}A^3 +R_3,
\end{align}
where
$R_3 = \sum_{n=4}^\infty\left(\frac25\right)^n\frac{\operatorname{tr}A^n}{n!} \ge 0$, since $e^x-(1+x+x^2/2+x^3/6)$ is positive for any $x\in\mathbb R$.
Then we can prove the inequality if we have
$$\frac25\operatorname{tr}A + \frac12\cdot \left(\frac25\right)^2\operatorname{tr}A^2+ \frac16\cdot \left(\frac25\right)^3\operatorname{tr}A^3\ge\!\!\!?\;0,$$
which can be simplified to
$$75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge\!\!\!?\;0.\tag{1}$$
Numerical results suggest that using the 3 first terms is enough to prove the inequality. Note that in the first case where $b\le1$, using the first term $\operatorname{tr}A$ is enough (and what we did in the first part is in fact proving $\operatorname{tr}A\ge0$), that's why that case is easy.
So,

*

*Why the case $b\ge1$ is difficult?
Because we have 2 more terms, $\operatorname{tr}A^2$ and $\operatorname{tr}A^3$, to deal with.


*Why $\frac25$ is important?
Because $\frac25$ gives the coefficients 75, 15, and 2, which makes $75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge0$.
A: Partial Hint :
With your work one can show that :
$$\left(\frac{a+1}{3-x-a+a} \right)\left(\frac{3-x-a+1}{3-x-a+x}\right)\left(\frac{x+1}{x+a} \right) \geqslant 1   $$
With $1\leq a\leq 2$ and $x\in[2-a,1]$
I cannot prove it but using Bernoulli's inequality we have :
$$ \left(\frac{a+1}{x+a} \right)^{\frac25}+\frac{1}{1+\frac{2}{5}\left(\frac{3-x-a+a}{3-x-a+1}-1\right)}+\left(\frac{x+1}{3-x-a+x} \right)^{\frac25}\geq 3$$
With $x\in[0,1]$ and $1\leq a\leq 1.5$
