Ta có:
$\dfrac{a}{a+b+c+d} < \dfrac{a}{a+b+c} < \dfrac{a}{a + c}$
$\dfrac{b}{a+b + c + d} < \dfrac{b}{b+c+d} < \dfrac{b}{b+d}$
$\dfrac{c}{a+b+c+d} <\dfrac{c}{c + d + a}<\dfrac{c}{a+c}$
$\dfrac{d}{a+b+c + d}<\dfrac{d}{d+a+b} <\dfrac{d}{b + d}$
Cộng vế theo vế ta được:
$\dfrac{a}{a+b+c+d}+ \dfrac{b}{a+b+c+d} + \dfrac{c}{a+b+c+d} + \dfrac{d}{a+b+c+d} < A <\dfrac{a}{a+c} + \dfrac{b}{b + d} +\dfrac{c}{a + c} + \dfrac{d}{b+d}$
$\Leftrightarrow 1 < A < 2\quad (đpcm)$