Solution 1. Denote the four terms by
$$A=\frac{(a-b)(a-c)}{a+b+c}, \quad B=\frac{(b-c)(b-d)}{b+c+d}, \quad C=\frac{(c-d)(c-a)}{c+d+a}, \quad D=\frac{(d-a)(d-b)}{d+a+b} .$$
The expression $2A$ splits into two summands as follows,
$$2A=A' + A'' \quad \text{ where } \quad A' = \frac{(a-c)^2}{a+b+c}, \quad A'' = \frac{(a-c)(a-2b+c)}{a+b+c}$$
this is easily verified. We analogously represent $2B = B' + B''$, $2C = C' + C''$, $2D = D' + D''$ and examine each of the sums $A'+B'+C'+D'$ and $A''+B''+C''+D''$ separately.
Write $s = a+b+c+d$; the denominators become $s-d$, $s-a$, $s-b$, $s-c$. By the Cauchy-Schwarz inequality,
$$\begin{aligned} & \left(\frac{|a-c|}{\sqrt{s-d}} \cdot \sqrt{s-d} + \frac{|b-d|}{\sqrt{s-a}} \cdot \sqrt{s-a} + \frac{|c-a|}{\sqrt{s-b}} \cdot \sqrt{s-b} + \frac{|d-b|}{\sqrt{s-c}} \cdot \sqrt{s-c}\right)^2 \\ & \quad \leq \left(\frac{(a-c)^2}{s-d} + \frac{(b-d)^2}{s-a} + \frac{(c-a)^2}{s-b} + \frac{(d-b)^2}{s-c}\right)(4s-s) = 3s\left(A'+B'+C'+D'\right) \end{aligned}$$
Hence
$$\begin{equation*} A'+B'+C'+D' \geq \frac{(2|a-c|+2|b-d|)^2}{3s} \geq \frac{16 \cdot |a-c| \cdot |b-d|}{3s} \tag{1} \end{equation*}$$
Next we estimate the absolute value of the other sum. We couple $A''$ with $C''$ to obtain
$$\begin{aligned} A''+C'' & = \frac{(a-c)(a+c-2b)}{s-d} + \frac{(c-a)(c+a-2d)}{s-b} \\ & = \frac{(a-c)(a+c-2b)(s-b) + (c-a)(c+a-2d)(s-d)}{(s-d)(s-b)} \\ & = \frac{(a-c)(-2b(s-b)-b(a+c)+2d(s-d)+d(a+c))}{s(a+c)+bd} \\ & = \frac{3(a-c)(d-b)(a+c)}{M}, \quad \text{ with } \quad M = s(a+c)+bd \end{aligned}$$
Hence by cyclic shift
$$B''+D'' = \frac{3(b-d)(a-c)(b+d)}{N}, \quad \text{ with } \quad N = s(b+d)+ca$$
Thus
$$\begin{equation*} A''+B''+C''+D'' = 3(a-c)(b-d)\left(\frac{b+d}{N} - \frac{a+c}{M}\right) = \frac{3(a-c)(b-d)W}{MN} \tag{2} \end{equation*}$$
where
$$\begin{equation*} W = (b+d)M - (a+c)N = bd(b+d) - ac(a+c) \tag{3} \end{equation*}$$
Note that
$$\begin{equation*} MN > ac(a+c) + bd(b+d) s \geq |W| \cdot s . \tag{4} \end{equation*}$$
Now (2) and (4) yield
$$\begin{equation*} \left|A''+B''+C''+D''\right| \leq \frac{3 \cdot |a-c| \cdot |b-d|}{s} . \tag{5} \end{equation*}$$
Combined with (1) this results in
$$\begin{aligned} & 2(A+B+C+D) = \left(A'+B'+C'+D'\right) + \left(A''+B''+C''+D''\right) \\ & \quad \geq \frac{16 \cdot |a-c| \cdot |b-d|}{3s} - \frac{3 \cdot |a-c| \cdot |b-d|}{s} = \frac{7 \cdot |a-c| \cdot |b-d|}{3(a+b+c+d)} \geq 0 \end{aligned}$$
This is the required inequality. From the last line we see that equality can be achieved only if either $a = c$ or $b = d$. Since we also need equality in (1), this implies that actually $a = c$ and $b = d$ must hold simultaneously, which is obviously also a sufficient condition.

Solution 2. We keep the notations $A$, $B$, $C$, $D$, $s$, and also $M$, $N$, $W$ from the preceding solution; the definitions of $M$, $N$, $W$ and relations (3), (4) in that solution did not depend on the foregoing considerations. Starting from
$$2A = \frac{(a-c)^2 + 3(a+c)(a-c)}{a+b+c} - 2a + 2c,$$
we get
$$\begin{aligned} & 2(A+C) = (a-c)^2\left(\frac{1}{s-d} + \frac{1}{s-b}\right) + 3(a+c)(a-c)\left(\frac{1}{s-d} - \frac{1}{s-b}\right) \\ & \quad = (a-c)^2 \frac{2s-b-d}{M} + 3(a+c)(a-c) \cdot \frac{d-b}{M} = \frac{p(a-c)^2 - 3(a+c)(a-c)(b-d)}{M} \end{aligned}$$
where $p = 2s-b-d = s+a+c$. Similarly, writing $q = s+b+d$ we have
$$2(B+D) = \frac{q(b-d)^2 - 3(b+d)(b-d)(c-a)}{N} ;$$
specific grouping of terms in the numerators has its aim. Note that $pq > 2s^2$. By adding the fractions expressing $2(A+C)$ and $2(B+D)$,
$$2(A+B+C+D) = \frac{p(a-c)^2}{M} + \frac{3(a-c)(b-d)W}{MN} + \frac{q(b-d)^2}{N}$$
with $W$ defined by (3).
Substitution $x = (a-c)/M$, $y = (b-d)/N$ brings the required inequality to the form
$$\begin{equation*} 2(A+B+C+D) = Mp x^2 + 3W x y + Nq y^2 \geq 0 . \tag{6} \end{equation*}$$
It will be enough to verify that the discriminant $\Delta = 9W^2 - 4MNpq$ of the quadratic trinomial $Mp t^2 + 3W t + Nq$ is negative; on setting $t = x/y$ one then gets (6). The first inequality in (4) together with $pq > 2s^2$ imply $4MNpq > 8s^3(ac(a+c)+bd(b+d))$. Since
$$(a+c)s^3 > (a+c)^4 \geq 4ac(a+c)^2 \quad \text{ and likewise } \quad (b+d)s^3 > 4bd(b+d)^2$$
the estimate continues as follows,
$$4MNpq > 8\left(4(ac)^2(a+c)^2 + 4(bd)^2(b+d)^2\right) > 32(bd(b+d)-ac(a+c))^2 = 32W^2 \geq 9W^2 .$$
Thus indeed $\Delta < 0$. The desired inequality (6) hence results. It becomes an equality if and only if $x = y = 0$; equivalently, if and only if $a = c$ and simultaneously $b = d$.

Solution 3.
$$\begin{gathered} (a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\ =((a-b)(a+b+d))((a-c)(a+c+d))(b+c+d)= \\ =\left(a^2+ad-b^2-bd\right)\left(a^2+ad-c^2-cd\right)(b+c+d)= \\ =\left(a^4+2a^3d-a^2b^2-a^2bd-a^2c^2-a^2cd+a^2d^2-ab^2d-abd^2-ac^2d-acd^2+b^2c^2+b^2cd+b c^2d+bcd^2\right)(b+c+d)= \\ a^4b+a^4c+a^4d+\left(b^3c^2+a^2d^3\right)-a^2c^3+\left(2a^3d^2-b^3a^2+c^3b^2\right)+ \\ +\left(b^3cd-c^3da-d^3ab\right)+\left(2a^3bd+c^3db-d^3ac\right)+\left(2a^3cd-b^3da+d^3bc\right) \\ +\left(-a^2b^2c+3b^2c^2d-2ac^2d^2\right)+\left(-2a^2b^2d+2bc^2d^2\right)+\left(-a^2bc^2-2a^2c^2d-2ab^2d^2+2b^2cd^2\right)+ \\ +\left(-2a^2bcd-ab^2cd-abc^2d-2abcd^2\right) \end{gathered}$$
Introducing the notation $S_{xyzw} = \sum_{cyc} a^x b^y c^z d^w$, one can write
$$\begin{gathered} \sum_{cyc}(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\ =S_{4100}+S_{4010}+S_{4001}+2S_{3200}-S_{3020}+2S_{3002}-S_{3110}+2S_{3101}+2S_{3011}-3S_{2120}-6S_{2111}= \\ +\left(S_{4100}+S_{4001}+\frac{1}{2}S_{3110}+\frac{1}{2}S_{3011}-3S_{2120}\right)+ \\ +\left(S_{4010}-S_{3020}-\frac{3}{2}S_{3110}+\frac{3}{2}S_{3011}+\frac{9}{16}S_{2210}+\frac{9}{16}S_{2201}-\frac{9}{8}S_{2111}\right)+ \\ +\frac{9}{16}\left(S_{3200}-S_{2210}-S_{2201}+S_{3002}\right)+\frac{23}{16}\left(S_{3200}-2S_{3101}+S_{3002}\right)+\frac{39}{8}\left(S_{3101}-S_{2111}\right), \end{gathered}$$
where the expressions
$$\begin{gathered} S_{4100}+S_{4001}+\frac{1}{2}S_{3110}+\frac{1}{2}S_{3011}-3S_{2120}=\sum_{cyc}\left(a^4b+bc^4+\frac{1}{2}a^3bc+\frac{1}{2}abc^3-3a^2bc^2\right), \\ S_{4010}-S_{3020}-\frac{3}{2}S_{3110}+\frac{3}{2}S_{3011}+\frac{9}{16}S_{2210}+\frac{9}{16}S_{2201}-\frac{9}{8}S_{2111}=\sum_{cyc} a^2c\left(a-c-\frac{3}{4}b+\frac{3}{4}d\right)^2, \\ S_{3200}-S_{2210}-S_{2201}+S_{3002}=\sum_{cyc} b^2\left(a^3-a^2c-ac^2+c^3\right)=\sum_{cyc} b^2(a+c)(a-c)^2, \\ S_{3200}-2S_{3101}+S_{3002}=\sum_{cyc} a^3(b-d)^2 \quad \text{ and } \quad S_{3101}-S_{2111}=\frac{1}{3} \sum_{cyc} bd\left(2a^3+c^3-3a^2c\right) \end{gathered}$$
are all nonnegative.