Prove $\left(\frac{a+1}{a+b}\right)^a+\left(\frac{b+1}{b+c}\right)^b+\left(\frac{c+1}{c+a}\right)^c \geqslant 3$ $a,b,c \geqslant 0,$$ a+b+c=3$, and $(a+b)(b+c)(c+a) \neq 0$ , prove

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

I try Bernouli's inequality but checking the case of $a>1$ and $0<a<1$ is truly complicated. I try to come up with some estimations (i.e., $x^x \geqslant \frac12 (1+x^2$)) but has not yet been successful. 

Yuri Negometyanov gave
$$\left(\frac{a+1}{a+b}\right)^a = \left(1 + \frac{b-1}{a+1}\right)^{-a} \ge 1 - a\cdot \frac{b-1}{a+1}
= 2 - b + \frac{b-1}{a+1}. \tag{1}$$
This follows from the Bernoulli inequality $(1+x)^r \ge 1 + rx$ for $x > -1$ and $r \le 0$.
Using (1), it suffices to prove that
$$\frac{b-1}{1+a}+ \frac{c-1}{1+b}+ \frac{a-1}{1+c}\ge 0. \tag{2}$$
We can prove (2) in another way as follows.
After clearing the denominators, it suffices to prove that
$$a^2b + b^2c + c^2a + a^2 + b^2 + c^2 - a - b - c - 3 \ge 0$$
or (since $a+b+c=3$)
$$a^2b + b^2c + c^2a - 2(ab+bc+ca) + 3 \ge 0.$$
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$a^2b + b^2c + c^2a \ge \frac{(ab + bc + ca)^2}{b + c + a} = \frac{(ab + bc + ca)^2}{3}.$$
Thus, we have
\begin{align}
a^2b + b^2c + c^2a - 2(ab+bc+ca) + 3 &\ge \frac{(ab + bc + ca)^2}{3} - 2(ab+bc+ca) + 3\\
&= \frac{1}{3}(ab + bc + ca - 3)^2\\
&\ge 0.
\end{align}
We are done.
