Certainly not the simplest way in terms of technical details, but I think it is very methodical. We want $(x_1 + x_3)(x_2 + x_4) \geq 4$, without loss of generality treat $a\leq b \leq c \leq d$ and we shall prove $(a+d)(b+c) \geq 4$. If we replace $b$ and $c$ with $x$, such that $\frac{1}{b+3} + \frac{1}{c+3} = \frac{2}{x+3}$, i.e. $x = \frac{2bc+3b+3c}{b+c+6}$, then the given condition is preserved, while $b+c\geq 2x$, as it is equivalent to $(b-c)^2 \geq 0$; it is also quick to check $b \leq x \leq c$. (Now we might think of doing the same with $a$ and $d$, but this is not allowed, as with the new values the ordering $a \leq b \leq c \leq d$ will not be valid anymore!) Hence it is enough to work with $b=c$, i.e. given $\frac{1}{a+3} + \frac{1}{d+3} + \frac{2}{b+3} = 1$ to show $b(a+d) \geq 2$. We have $b = \frac{9-ad}{ad+2a+2d+3}$ and have to show $\frac{(a+d)(9-ad)}{ad+2a+2d+3}\geq 2$. The inequality $b \geq a$ is equivalent to $a \leq \frac{3}{d+2}$, while $b\leq d$ is equivalent to $a \geq \frac{3}{d} - 2$. We wish to prove \[ da^2 + (d^2+2d-5)a - 5d + 6 \leq 0. \]This quadratic function on $a$ is with a positive leading coefficient (if $d=0$, then $a=b=c=0$, impossible) and so it is non-positive for $a\in \left[\frac{3}{d}-2, \frac{3}{d+2} \right]$ if and only if it is non-positive for $a = \frac{3}{d}-2$ and for $a = \frac{3}{d+2}$. For the former, it is equal to $-\frac{2(d-1)^2(d+3)}{d} \leq 0$, while for the latter it is $-\frac{2(d-1)^2(d+3)}{(d+2)^2} \leq 0$ and so we are done. Equality in the latter two is reached only when $d=1$, then the above interval for $a$ yields $a=1$ and the ordering $a\leq b \leq c \leq d$ concludes that equality holds if and only if $a=b=c=d=1$.

wlog, let $a\geq b,d\geq c$. Then there exists $x\geq y,w\geq z$ such that $\left(a,b,c,d\right) = \left(\frac{3x}{y+z+w},\dfrac{3y}{z+w+x},\dfrac{3z}{w+x+y},\dfrac{3w}{x+y+z}\right)$. We will prove $\left(a+c\right)\left(b+d\right)\geq 4$, which is equivalent to $ \left(\frac{3x}{y+z+w}+\dfrac{3z}{w+x+y}\right)\left(\dfrac{3y}{z+w+x}+\dfrac{3w}{x+y+z}\right)$. Notice that by Schwarz, $\frac{3x}{y+z+w}+\dfrac{3z}{w+x+y}\geq \dfrac{3\left(x+z\right)^2}{2xz + \left(x+z\right)\left(y+w\right)}, \dfrac{3y}{z+w+x}+\dfrac{3w}{x+y+z}\geq \dfrac{3\left(y+w\right)^2}{2yw+\left(x+z\right)\left(y+w\right)}$. Hence, we only need to prove $4\left(2xz+S\right)\left(2yt + S\right)\leq 9S^2$, where $S = \left(x+z\right)\left(y+w\right)$. However, $S-2(xz+yt) = \left(x+z\right)\left(y+w\right)-2(xz+yt) = (x-w)(y-z) + (w-z)(x-y)\geq 0$, hence, $2(xz + yt) \leq S$ and $16xyzt \leq S^2$. Thus, $4(2xz + S)(2yt + S)\leq S^2+4S^2 + 4S^2 = 9S^2$, completing the problem.

We do this via smoothing. Notice that $x_1x_2+x_2x_3+x_3x_4+x_4x_1=(x_1+x_3)(x_2+x_4)$, and furthermore note that if $a\le b\le c\le d$ WLOG then $(x_1+x_3)(x_2+x_4)=(a+d)(b+c)$ maximises the product, which is pretty clear. Let $w=\frac{1}{a+3}$, etc, so that the condition reduces down to $w+x+y+z=1$ and the inequality itself becomes $\left(\frac{1}{w}+\frac{1}z-6\right)\left(\frac 1 x+\frac 1 y-6\right)\ge4$. We now do the following steps: Smooth $x$ and $y$ together so that we may assume $x=y$. Smooth $w$ and $z$ together as close to each other without breaking the condition $w\ge x\ge y\ge z$ so that we can now assume three values equal Calculate the minimum of our resulting function with three equal values and show it is equal to $4$. The first two are easy; noticing that $f(x)=\frac{1}{x}$ is convex by Karamata we may smooth $x$ and $y$ together, and same with $w$ and $z$. So assume we have three equal values. The expression now reduces down to $2\left(\frac{1}{x}-3\right)\left(\frac{1}{1-3x}+\frac{1}{x}-6\right)$. At this point simple calculus methods suffice, albeit with a lot of algebraic gruntwork. To move past that, we expand and get that the expression being $f(x)=2\left(18-\frac{8}{x}+\frac{1}{x^2}\right)$. Notice however that $f(x)$’s minimum in $(0,\frac{1}{3}]$ is the same as the minimum of $g(x)=f(\frac 1 x)$ in $[3,\infty)$. Notice that $g(x)$ is just $2x^2-16x+36$, which has its minimum at its vertex $x=-\frac{-16}{2\cdot2}=4$ with value $g(4)=2(4-4)^2+4=4$, as desired. Hence $(a+d)(b+c)\ge4$ always as desired.