$\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + .... + \dfrac{1}{300}$

$= (\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + .... + \dfrac{1}{200}) + (\dfrac{1}{201} + \dfrac{1}{202} + ... + \dfrac{1}{300})$

$\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + .... + \dfrac{1}{200} > \dfrac{1}{200} + \dfrac{1}{200} + \dfrac{1}{200} + .... + \dfrac{1}{200}$

$\dfrac{1}{201} + \dfrac{1}{202} + \dfrac{1}{203} + .... + \dfrac{1}{300} > \dfrac{1}{300} + \dfrac{1}{300} + \dfrac{1}{300} + .... + \dfrac{1}{300}$

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

$⇒ A > \dfrac{2}{3}$ ($đpcm$)

Ta đặt `A=1/101+1/102+1/103+...+1/300`

`⇒A=(1/101+1/102+...+1/200)+(1/201+1/202+...+1/300)`

`⇒A>(1/200+1/200+...+1/200)+(1/201+1/202+...+1/300)`

`⇒A>(\underbrace{1/200+1/200+...+1/200}_{\text{100 chữ số}})+(\underbrace{1/300+1/300+...+1/300}_{\text{100 chữ số}})`

`⇒A>1/200 . 100+1/300 . 100`

`⇒A>100/200+100/300`

`⇒A>1/2+1/3`

`⇒A>5/6>4/6=2/3`

`⇒A>2/3`