Here's a proof.
Using Lagrange's identity leads to
$$\sum_{cyc}\frac{x}{y+z}=\frac32+\frac12\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \qquad (*)$$
We get this by letting $a = \sqrt{x+y}$, $b = \sqrt{y+z}$, $c = \sqrt{z+x}$,
$d = \frac{1}{\sqrt{x+y}}$, $e = \frac{1}{\sqrt{y+z}}$, $f = \frac{1}{\sqrt{z+x}}$, and by Lagrange's identity,
$$(a^2+b^2+c^2)(d^2+e^2+f^2) = (ad+be+cf)^2 + (ae-bd)^2+(af-cd)^2+(bf-ce)^2$$
which gives
$$2 (x+y+z)(\sum_{cyc}\frac{1}{y+z}) = 9 + \sum_{cyc} \left(\frac{\sqrt{x+y}}{\sqrt{y+z}} - \frac{\sqrt{y+z}}{\sqrt{x+y}}\right)^2$$
or
$$3 +\sum_{cyc}\frac{x}{y+z} = \frac92 + \frac12\sum_{cyc} \frac{\left({x+y} - (y+z)\right)^2}{({y+z})(x+y)} $$
which is the desired equation $(*)$.
Squaring both sides of the inequality then gives
$$
\tag{1} \frac16 \left[ \sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \right ]^2 + \sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geq \frac{(y-z)^2}{xy + yz + xz}
$$
We will follow two paths for separate cases.
Path 1:
Omitting the square term it suffices to prove
$$\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant \frac{(y-z)^2}{xy + yz + xz}$$
Clearing denominators, we obtain
$$
(x-y)^2 (x+y) + (y-z)^2 (y+z) + (z-x)^2 (z+x) \geq \frac{(y-z)^2}{xy + yz + xz} (x+y) (y+z) (z+x)
$$
Using $(x+y)(y+z)(x+z) = (x+y+z)(xy + yz+ xz) - x y z$
it suffices to show
$$
(x-y)^2 (x+y) + (y-z)^2 (y+z) + (z-x)^2 (z+x) \geq {(y-z)^2} (x+y+z)
$$
or
$$
(x-y)^2 (x+y) + (z-x)^2 (z+x) \geq (y-z)^2 x
$$
Since
$$
(y-z)^2 = (y-x + x- z)^2 = (y-x)^2 + (x- z)^2 + 2 (y-x)(x-z)
$$
this translates into
$$
(y-x)^2 y + (x-z)^2 z \geq 2 x(y-x)(x-z)
$$
For the two cases $y\geq x ; z\geq x $ and $y\leq x ; z\leq x $ the RHS $\leq 0$ so we are done. For the other two cases, by symmetry, it remains to show the case $y> x ; z < x $.
Rearranging terms, we can also write
$$
(y-x)^3 - (x-z)^3 +x ((y - x) + (z-x))^2 \geq 0
$$
This holds true at least for $(y-x)^3 \geq (x-z)^3$ or $y+z\geq 2 x$.
So the proof is complete other than for the case $y+z < 2x$ and [ $y> x ; z < x $ or $z> x ; y < x $ ].
Path 2.
For the remaining case $y+z < 2 x $ and [$y> x ; z < x$ or $z> x ; y < x $] we will follow a different path. Again, by symmetry, we must inspect only $y+z < 2x$ and $y> x > z$.
A remark up front: In the following, some high order polynomials of one variable have to be inspected. MATLAB is used, also for plotting behaviours of these polynomials. There is no case in spending effort for further analytical work on polynomials where their behaviour is obvious. Still, what follows contains some "ugly" parts.
In the squared version (1) of the inequality, we can use a further inequality which has been proved here:
$$\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2}$$
So it suffices to prove
$$
\frac16 \left[ \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2} \right ]^2 + \sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geq \frac{(y-z)^2}{xy + yz + xz}
$$
Some numerical inspection shows immediately that the first term cannot be ommitted.
Clearing some denominators gives
$$
\tag{2} \frac16 \left[ \frac{27}{8} \right ]^2 \frac{(y-z)^4 (xy + yz + xz) (x+y)(y+z) (z+x)}{(x+y+z)^4} + \\
(xy + yz + xz) \sum_{cyc} {(x-y)^2}{(x+y)} - (y-z)^2 (x+y)(y+z) (z+x) \geqslant 0 \quad
$$
By homogeneity, we set $y=1+z$.
The condition $y+z < 2x$ then translates into $1+2z < 2x$, hence we further set $x = z + (1 +q)/2$ where $0\leq q \leq 1$ since also $x = z + (1 +q)/2 < y = 1 +z$.
Inserting $y=1+z$ and $x = z + (1 +q)/2$ into (2) is straightforward, the result is lengthy (not displayed here).
Let us start with focussing on the first term in (2), calling that fraction $F$:
$$
F= \frac{(y-z)^4 (xy + yz + xz) (x+y)(y+z) (z+x)}{(x+y+z)^4}
$$
With the setting $y=1+z$ this can be simplified to
$$
F = \frac{(xy + yz + xz) (x+y)(y+z) (z+x)}{(x+y+z)^4}
$$
Since by the settings $y=1+z$ and $x = z + (1 +q)/2$ both x and y are linear in z, the numerator of $F$ is of fifth order in $z$, whereas the denominator is of fourth order in $z$. In leading order, the whole term is therefore of first order in $z$ and will therefore rise with $z$ for large enough $z$. This motivates to show that indeed
$$F(q,z) \geq F(q,z=0) = 2 \frac{(q+1)^2}{(q+3)^2}$$
for all $z$ and $q$.
Showing that directly requires a condition
$$
G = (xy + yz + xz) (x+y)(y+z) (z+x) (q+3)^2 - 2 (q+1)^2 (x+y+z)^4 \geq 0
$$
Inserting $y=1+z$ and $x = z + (1 +q)/2$ into $G$, and expanding the brackets, gives a very lengthy expression which however contains only positive terms, so the condition is proved immediately:
$$
G = (z(2q^6z + 2q^6 + 22q^5z^2 + 51q^5z + 22q^5 + 80q^4z^3 + 358q^4z^2 + 357q^4z + 100q^4 + 96q^3z^4 + 960q^3z^3 + 1828q^3z^2 + 1206q^3z + 260q^3 + 864q^2z^4 + 3672q^2z^3 + 4788q^2z^2 + 2484q^2z + 450q^2 + 2592qz^4 + 7344qz^3 + 7398qz^2 + 3159qz + 486q + 2592z^4 + 5832z^3 + 4806z^2 + 1701z + 216))/4 \geq 0
$$
Hence it suffices, instead of (2), to prove the following:
$$
\tag{3} \frac16 \left[ \frac{27}{8} \right ]^2 2 \frac{(q+1)^2}{(q+3)^2} +
(xy + yz + xz) \sum_{cyc} {(x-y)^2}{(x+y)} \\ - (x+y)(y+z) (z+x) \geqslant 0 \quad
$$
After insertion of $y=1+z$ and $x = z + (1 +q)/2$, the factors $(x-y)^2$ in the cyclic sum will not be functions of $z$. Hence,
the LHS is a third order expression in $z$ with leading (in $z^3$ ) term $( 1 + 3 q^2) z^3$, so for large enough $z$ it is rising with $z$. A remarkable feature of this expression is that for the considered range $0\leq q \leq 1$ it is actually monotonously rising for all $z$. To see this, consider whether there are points with zero slope. The first derivative of the expression with respect to $z$ is
$$
q^4/4 + (7q^3z)/2 + (7q^3)/4 + 9q^2z^2 + 9q^2z + (5q^2)/4 - (7qz)/2 - (7q)/4 + 3z^2 + 3z + 1/2
$$
Equating this to zero gives
$$
z_{1} =
-(2((13q^6)/4 + (17q^4)/2 + (133q^2)/4 + 3)^{(1/2)} - 7q + 18q^2 + 7q^3 + 6)/(36q^2 + 12)\\
z_{2} =
-(-2((13q^6)/4 + (17q^4)/2 + (133q^2)/4 + 3)^{(1/2)} - 7q + 18q^2 + 7q^3 + 6)/(36q^2 + 12)
$$
One now shows that in the range $0\leq q \leq 1$, there are only negative solutions $z_{1,2}$. So we will have no zero slopes for $z\geq 0$. Since the polynomials are (roots of) sixth order in q, we investigate the following figures:
So monotonicity (rising with $z$) is established for all $q$.
Hence, to show the inequality it suffices to inspect (3) at the smallest $z=0$. This gives
$$
((q + 1)(8q^6 + 88q^5 + 336q^4 + 432q^3 - 216q^2 - 405q + 243))/(64(q + 3)^3) \geqslant 0 \quad
$$
or
$$
\tag{4} 8q^6 + 88q^5 + 336q^4 + 432q^3 - 216q^2 - 405q + 243 \geqslant 0 \quad
$$
An even stronger requirement is
$$
h(q) =
432q^3 - 216q^2 - 405q + 243 \geq 0
$$
In the considered range $0\leq q \leq 1$, $h(q)$ has a minimum which is obtained by taking the first derivative,
$$
1296 q^2 - 432 q - 405
$$
and equating to zero, which gives $q = 3/4$, and the above $h(q)$ then gives
$$
h(q = 3/4) = 0
$$
This establishes the inequality. $ \qquad \Box$
$$\mathbf{\color{green}{New\ proof,\ version\ of\ 02.07.18}}$$
$$\mathbf{\color{brown}{Using\ inequality}}$$
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
Since $e^x\geq 1+x$ for all $x\in \mathbb{R}$, by letting $x=-\log u$ we have that for any $u >0$, $\log u\geq 1-\frac{1}{u}$, and so $u\log u>u-1$ for any $u>0$. Combining these two inequalities, it follows that $$\sqrt{x^x y^y}= \exp\left(\frac{x\log x}{2}+\frac{y\log y}{2}\right)\geq \exp\left(\frac{x+y}{2}-1\right)\geq \frac{x+y}{2}.$$ Hence $$\frac{x^x}{x+y}+\frac{y^y}{y+z}+\frac{z^z}{z+x}\geq \frac{1}{2} \left(\sqrt{\frac{x^x}{y^y}}+\sqrt{\frac{y^y}{z^z}}+\sqrt{\frac{z^z}{x^x}}\right),$$ and the result follows from the AM-GM inequality.
Remark: By combining $e^x\geq 1+x$ for $x\in\mathbb{R}$ and $u\log u\geq u-1$ for $u>0$ as above, we obtain the following "reverse exponential AM-GM inequality": $$\sqrt[n]{x_{1}^{x_{1}}x_{2}^{x_{2}}\cdots x_{n}^{x_{n}}}\geq \frac{x_1+x_2+\cdots+x_n}{n}.$$