Giải thích các bước giải:
Ta có : $a<a+b+c$
$\to \dfrac{a}{a+b+c+d}<\dfrac{a}{a+b+c}<\dfrac{a+d}{a+b+c+d}$
Tương tự ta có :
$\dfrac{b}{a+b+c+d}<\dfrac{b}{b+c+d}<\dfrac{b+a}{a+b+c+d}$
$\dfrac{c}{a+b+c+d}<\dfrac{c}{c+d+a}<\dfrac{c+b}{a+b+c+d}$
$\dfrac{d}{a+b+c+d}<\dfrac{d}{d+a+b}<\dfrac{d+c}{a+b+c+d}$
$\to \dfrac{a}{a+b+c+d}+ \dfrac{b}{a+b+c+d}+ \dfrac{c}{a+b+c+d}+ \dfrac{d}{a+b+c+d}<N\to N>1$
Lại có : $N<\dfrac{a+d}{a+b+c+d}+\dfrac{b+a}{a+b+c+d}+\dfrac{c+b}{a+b+c+d}+\dfrac{d+c}{a+b+c+d}=2$
$\to 1<N<2\to N\notin Z$
