This is Nesbitt inequality for 4 variables. Call the expression A

1

We called $B=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}$ $C=\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}+\frac{b}{a+b}$ Now by AM-GM we get $A+B=\frac{a+b}{b+c}+\frac{b+c}{c+d}+\frac{c+d}{d+a}+\frac{d+a}{a+b}\geq 4$ Since $a,b,c,d$ are non-negatives, we have $A+C=\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{a+c}{d+a}+\frac{b+d}{a+b}+\frac{a+c}{c+b}+\frac{a+c}{a+d}+\frac{b+d}{c+d}+\frac{b+d}{a+b}\geq \frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}=4$ Finally $2A+B+C\geq 8 \Rightarrow A\geq 2$ The equality holds when $a=b=c=d$

2

Since the first solution is too complicated, here's an another solution. We called $G=a(b+c)+b(c+d)+c(a+d)+d(a+b)=(a+b)(c+d)+(a+d)(b+c)$ Now applying Cauchy-Schwarz inequality we get $A.G\geq (a+b+c+d)^2$ Clearly see that $G\leq \frac{(a+b+c+d)^2}{2}$ by AM-GM $\therefore A\geq 2$ The equality holds when $a=b=c=d$