Ta có : 

S = \(\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)

Đặt :

\(A=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(A=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Ta thấy :

\(\dfrac{1}{13}< \dfrac{1}{12};\dfrac{1}{14}< \dfrac{1}{12};\dfrac{1}{15}< \dfrac{1}{12}\)

\(\dfrac{1}{61}< \dfrac{1}{60};\dfrac{1}{62}< \dfrac{1}{60};\dfrac{1}{63}< \dfrac{1}{60}\)

\(\Rightarrow A< \dfrac{1}{5}+\left(\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}\right)+\left(\dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}\right)\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)

\(\Rightarrow A< \dfrac{1}{2}\rightarrowđpcm\)

~ Chúc bn học tốt ~

TA có:

1/12>1/13

1/12>1/14

1/12>1/15

=>1/12.3=1/4>1/13+1/14+1/15

1/60>1/61

1/60>1/62

1/60>1/63

=>1/60.3=1/20>1/61+1/62+1/63

=>1/5+1/4+1/20> 1/5+1/13+1/14+1/15+1/61+1/62+1/63

=>1/2> 1/5+1/13+1/14+1/15+1/61+1/62+1/63

Ta có: 

\(\frac{1}{5}=\frac{1}{5}\)

\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}

Ta có: \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)

\(đpcm\Leftrightarrow\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}