Proving $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2$ If $a,b,c,d \in \mathbb{N}$

Question

Given pairwise distinct $a,b,c,d \in \mathbb{N}$, prove that $$E=2$$ if $E=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}$ is an integer.

My effort: We have: $$\begin{aligned} & E=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a} \\ & F=\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{a+d} \end{aligned}$$

WLOG let $a>b>c>d$, so $$\begin{aligned} & \frac{b}{a}<1 \\ & \Rightarrow \quad \frac{1}{1+\frac{b}{a}}>\frac{1}{2} \Rightarrow \frac{a}{a+b}>\frac{1}{2} \end{aligned}$$ Like-wise:$$\frac{b}{b+c}>\frac{1}{2},\:\frac{c}{c+d}>\frac{1}{2} \Rightarrow E>1.5 \Rightarrow F<2.5 ----(1)$$ Which gives two possibilities $E=F=2$ or $E=3,F=1$. How to rule out the second possibility?

Answer 1

$$E>\frac a{a+b+c+d}+\frac b{a+b+c+d}+\frac c{a+b+c+d}+\frac d{a+b+c+d}=1 $$

Similarly, $F=\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}>1$.

Since $E+F=(\frac a{a+b}+\frac b{a+b})+(\frac b{b+c}+\frac c{b+c})+(\frac c{c+d}+\frac d{c+d})+(\frac d{d+a}+\frac a{d+a})=4$, we have $E=4-F<3$.
Since $E$ is an integer and $1<E<3$, we have $E=2$.

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The proof above does not require any one of $a,b,c,d$ be integer. What is needed is $a,b,c,d>0$.