$\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + .... + \dfrac{1}{300}$
$= (\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + .... + \dfrac{1}{200}) + (\dfrac{1}{201} + \dfrac{1}{202} + ... + \dfrac{1}{300})$
$\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + .... + \dfrac{1}{200} > \dfrac{1}{200} + \dfrac{1}{200} + \dfrac{1}{200} + .... + \dfrac{1}{200}$
$\dfrac{1}{201} + \dfrac{1}{202} + \dfrac{1}{203} + .... + \dfrac{1}{300} > \dfrac{1}{300} + \dfrac{1}{300} + \dfrac{1}{300} + .... + \dfrac{1}{300}$
$⇒ (\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + .... + \dfrac{1}{200}) + (\dfrac{1}{201} + \dfrac{1}{202} + ... + \dfrac{1}{300}) > (\dfrac{1}{200} + \dfrac{1}{200} + \dfrac{1}{200} + .... + \dfrac{1}{200}) + (\dfrac{1}{300} + \dfrac{1}{300} + \dfrac{1}{300} + .... + \dfrac{1}{300}) = \dfrac{1}{200} . 100 + \dfrac{1}{300} . 100 = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6} > \dfrac{4}{6} = \dfrac{2}{3}$
$⇒ A > \dfrac{2}{3}$ ($đpcm$)
