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

\(\Rightarrow S< \frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)

\(\Rightarrow S< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}\)

\(\Rightarrow S< \frac{1}{2}\)

gọi đó là A đi.

Ta có:

1/13+1/14+1/14< 1/12+1/12+1/12=3/12=1/4

1/61+1/62+1/63< 1/60+1/60+1/60=3/60=1/20

=> 1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/5+1/4+1/20=1/2

=>A< 1/2 (ĐPCM)

S=1/5=(1/13+1/14+1/15)+(1/61+1/62+1/63)

suy ra S<1/5+1/12x3+1/60x3

S<1/5+1/4+1/20

=>S<1/2

Lời giải:

Ta có:

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

\(\Rightarrow \frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{3}{12}=\frac{1}{4}\)

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

\(\Rightarrow \frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{3}{60}=\frac{1}{20}\)

Do đó:

\(A< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{9}{20}+\frac{1}{20}\)

\(\Leftrightarrow A< \frac{1}{2}\) (đpcm)

Đặt biểu thức bằng A:

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

Ta thấy: \(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< 3.\dfrac{1}{61}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< 3.\dfrac{1}{61}\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{3}{31}+\dfrac{3}{61}< \dfrac{1}{2}\left(đpcm\right)\)