$$\mathbf{\color{green}{New\ proof,\ version\ of\ 02.07.18}}$$
$$\mathbf{\color{brown}{Using\ inequality}}$$
Let us consider the inequality
$$(1+t)^{-\alpha}\geq1-\alpha t\quad \text{ for } t>-1,\quad \alpha>0,\tag1$$
which works due to the Maclauin series of
$$(1+t)^{-\alpha} = 1 - \alpha t+\dfrac12(-\alpha)(-\alpha-1)t^2+\dfrac16(-\alpha)(-\alpha-1)(\alpha-2)t^3+\dots$$
Using of the inequality $(1)$ in the form of
$$\left(\dfrac{a+1}{a+b}\right)^a=\left(\dfrac{a+b}{a+1}\right)^{-a}= \left(1+\dfrac{b-1}{a+1}\right)^{-a}\ge1-\dfrac{a(b-1)}{1+a}=2-b+\dfrac{b-1}{1+a}\tag2$$
is correct, because for the issue conditions
$$a>0,\ b>0,\ c>0,\quad a+b+c = 3\tag3$$
$$\dfrac{b-1}{a+1} > -1.$$
$$\mathbf{\color{brown}{Task\ transformation}}$$
Using the inequality $(2)$ with the constraints $(3),$ easily to get
$$\left(\dfrac{a+1}{a+b}\right)^a + \left(\dfrac{b+1}{b+c}\right)^b + \left(\dfrac{c+1}{c+a}\right)^c \geq 3+ \dfrac{b-1}{1+a}+ \dfrac{c-1}{1+b}+
\dfrac{a-1}{1+c},$$
so it can be proved the stronger inequality than the issue one:
$$\dfrac{b-1}{1+a}+ \dfrac{c-1}{1+b}+ \dfrac{a-1}{1+c}\ge0,$$
or
$$(b^2-1)(1+c)+(c^2-1)(1+a)+(a^2-1)(1+b)\ge0.\tag4$$
The least value of $LSH(4)$ can be achieved in the stationary points or on the edges of the area $(3).$
$$\mathbf{\color{brown}{Stationary\ points}}$$
The stationary points of the function can be found using the Lagrange multiplyers method, as the stationary points of the function
$$f(a,b,c,\lambda)=(b^2-1)(1+c)+(c^2-1)(1+a)+(a^2-1)(1+b)+\lambda(a+b+c-3),$$
by the solving of the systen $f'_a = f'_b = f'_c = f'_\lambda=0,$ or
\begin{cases}
c^2-1+2a(1+b)+\lambda=0\\
a^2-1+2b(1+c)+\lambda=0\\
b^2-1+2c(1+a)+\lambda=0\\
a+b+c=3.\tag5
\end{cases}
Summation of $(5.1)-(5.3)$ gives $\lambda=-4,$ then
\begin{cases}
c^2+2a(1+b)=a^2+2b(1+c)=b^2+2c(1+a)=5\\
a+b+c=3,
\end{cases}
\begin{cases}
c^2+2a(4-a-c)=a^2+2(3-a-c)(1+c)=5\\
a+b+c=3,
\end{cases}
\begin{cases}
3c^2=3a^2-10a+4c+6\\
3a^2-10a+4c+6+6a(4-a-c)=15\\
a+b+c=3,
\end{cases}
\begin{cases}
(4-6a)c=3a^2-14a+9\\
3(3a^2-14a+9)^2=(4-6a)^2(3a^2-10a+6)+4(4-6a)(3a^2-14a+9)\\
a+b+c=3,
\end{cases}
\begin{cases}
(a-1)(27a^3-81a^2+45a+1)=0\\
c=\dfrac{3a^2-14a+9}{4-6a}\\
a+b+c=3,\tag6
\end{cases}
The root $a=1$ leads to the solution
$$a=b=c=1,\quad f(a,b,c,\lambda)=0.\tag7$$
Taking in account the task symmetry by the variables $a,b,c,$ the roots of cubic part of $(6.1)$ are the values of $a,b,c.$ At the same time, production of this roots is negative due to the Vieta theorem. This means that the solution $(7)$ is the single one, which satisfies conditions $(3).$
$$\mathbf{\color{brown}{The\ edges}}$$
To analyze the edges of area let us consider the case $a\to 0.$ It leads to the inequality in one variable
$$(b^2-1)(2-b)+(3-b)^2-b-2\ge0,$$
or
$$3-b(b-2)(b-3)\ge0.\tag8$$
Taking in account that for $b\in[2,3]$ $LSH(8)\ge3,$ and for $b\in(0,2)$
$$LSH(8)=3-b(2-b)(3-b)\ge3-\dfrac{(5-b)^3}9\ge0,$$
the inequality $(8)$ holds in the edges.
$$\mathbf{\color{green}{Proved}}$$
$$\mathbf{\color{black}{Old\ proof}}$$
$$\mathbf{\color{brown}{Task\ transformation}}$$
First, we should consider the inequality within the area.
Using evident inequality
$$\dfrac1{1-t}\geq1+t\quad \text{ for } t\in\left(-1,1\right)\qquad(1)$$
and AM-GM for $a,b,c>0,$ one can get:
$$\dfrac{a+1}{a+b}=\dfrac{a+1}{3-c}=\dfrac12\,\dfrac{a+1}{1-\dfrac{c-1}2}\geq\dfrac{a+1}2\left(1+\dfrac{c-1}2\right) = \dfrac{a+1}2\,\dfrac{c+1}2\geq\sqrt{ac},$$
$$\left(\dfrac{a+1}{a+b}\right)^a + \left(\dfrac{b+1}{b+c}\right)^b +
\left(\dfrac{c+1}{c+a}\right)^c \geq (ac)^{a/2} + (ba)^{b/2} +(cb)^{c/2}\\
=\exp\left(\frac{a}2\,\ln(ac)\right) + \exp\left(\frac{b}2\,\ln(ba)\right) +
\exp\left(\frac{c}2\,\ln(cb)\right)\\
\geq 3\exp\left(\frac{a}6\,\ln(ac) + \frac{b}6\,\ln(ba)+\frac{c}6\,\ln(cb)\exp\right)\\
=3\exp\left(\frac{a+b}6\ln{a}+\frac{b+c}6\,\ln{b}+\frac{c+a}6\,\ln{c}\right) \\
=3\exp\left(\frac{3-c}6\,\ln{a}+\frac{3-a}6\,\ln{b}+\frac{3-b}6\, \ln{c}\right),$$
and that gives the possibility to prove the inequality
$$(3-c)\ln{a}+(3-a)\ln{b}+(3-b)\ln{c}\geq0\tag2$$
for $a,b,c>0,\ a+b+c=3.$
$$\mathbf{\color{brown}{Stationary\ points}}$$
To do this, it suffices to find the least value of the function
$$f(a,b) = (a+b)\ln{a}+(3-a)\ln{b}+(3-b)\ln(3-a-b).\tag3$$
The stationary points of the function can be found by equating to zero the partial derivatives $f'_a$ and $f'_b,$ or
$$\begin{cases}
\ln{a}+\dfrac{a+b}a -\ln{b} - \dfrac{3-b}{3-a-b} = 0\\[4pt]
\ln{a}+\dfrac{3-a}b - \ln(3-a-b) - \dfrac{3-b}{3-a-b} =0,
\end{cases}$$
$$\begin{cases}
\ln{\dfrac{a}b} + \dfrac{a}{b} - \dfrac{a}{3-a-b} = 0\\[4pt]
\ln{\dfrac{a}{3-a-b}} + \dfrac{3-a-b}{b} - \dfrac{a}{3-a-b} = 0,\\
\end{cases}$$
or
$$\begin{cases}
\ln{xy} - x + xy =0\\[4pt]
\ln{x} -x + y =0\\[4pt]
x=\dfrac{a}{3-a-b}\\[4pt]
y=\dfrac{3-a-b}{b},
\end{cases}$$
$$\ln{y}+y(x-1)=0,\quad y=x-\ln{x},$$
$$\ln(x-\ln{x})+(x-1)(x-\ln{x})=0,\tag4$$
with single solution
$$x=y=a=b=1,$$
which corresponds to the maximum $f(a,b)$
$$f_m=0.$$
$$\mathbf{\color{brown}{The\ edges}}$$
To analyze the edges of area let us consider the case $a=0.$ It leads to the inequality in
one variable
$$\left(\dfrac{b+1}3\right)^b+\left(\dfrac{4-b}{3-b}\right)^{3-b}\geq 2,\tag5$$
which is satisfied for $0<b<3.$
Thus,
$$\boxed{\left(\dfrac{a+1}{a+b}\right)^a + \left(\dfrac{b+1}{b+c}\right)^b +
\left(\dfrac{c+1}{c+a}\right)^c \geq 3}.$$
Best Answer
Proceeding along your approach:
If $S \ge \frac{5}{2}$, it is true.
If $0 \le S < \frac{5}{2}$, since $S^2 \ge 4P$, we have \begin{align} &P{S}^{2}+2{S}^{3}+20PS-71{S}^{2}+228P+52S-172\\ =\ & (S^2+20S+228)P+2S^3-71S^2+52S-172 \\ \le\ & (S^2+20S+228)\frac{S^2}{4}+2S^3-71S^2+52S-172\\ =\ & \frac{1}{4}S^4+7S^3-14S^2+52S-172 \\ <\ & 0 \end{align} and thus it suffices to prove that $$\frac{3}{4} ( \frac{1}{4}S^4+7S^3-14S^2+52S-172 ) ^{2}+48 ( 6{S}^{3}-13{S}^{2}+8S-28 ) ( S-4 ) ^{2} \ge 0$$ that is $$\frac{3}{64}(S^4+40S^3-120S^2+96S+16)(S^2+20S+228)(S-2)^2 \ge 0$$ which is true.
We are done.