Ta có: \(\frac{1}{5}\)>\(\frac{1}{14}\)

\(\frac{1}{14}\)=\(\frac{1}{14}\)

\(\frac{1}{28}\)<\(\frac{1}{14}\)

...

\(\frac{1}{97}< \frac{1}{14}\)

=>Cả dãy số < \(\frac{1}{14}.7\)<\(\frac{1}{2}\)

Hai dãy số trên chưa hai bđt ngược chiều: 1/5 > 1/14 và 1/28<1/14 ..... thì làm sao mà cộng theo vế để suy ra như vậy đc?

\(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}< \frac{1}{2}\)

\(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}=0,36833......\)

mà \(\frac{1}{2}=0,5\)

\(0,36833..< 0,5\)

Vậy \(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}< \frac{1}{2}\)

Đặt \(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}=A\)

Ta có : \(A=\frac{1}{5}+\left(\frac{1}{14}+\frac{1}{28}+\frac{1}{44}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{97}\right)\)

\(A< \frac{1}{5}\left(\frac{1}{14.3}\right)+\left(\frac{1}{61.3}\right)\)

\(A< \frac{1}{5}+\frac{3}{14}+\frac{3}{61}\)

\(A< \frac{1}{5}+\frac{3}{12}+\frac{1}{20}\)

\(A< \frac{1}{2}\left(ĐPM\right)\).

Ta có: \(\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}.\)

\(=\frac{1}{5}+\left(\frac{1}{14}+\frac{1}{31}+\frac{1}{44}\right)+\left(\frac{1}{61}+\frac{1}{84}+\frac{1}{96}\right)\)

Ta thấy \(\frac{1}{14}< \frac{1}{12}\)

            \(\frac{1}{31}< \frac{1}{12}\)

            \(\frac{1}{44}< \frac{1}{12}\)

\(=>\frac{1}{14}+\frac{1}{31}+\frac{1}{44}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}\)

\(=>\frac{1}{14}+\frac{1}{31}+\frac{1}{44}< \frac{1}{12}.3\left(1\right)\)

Ta lại thấy \(\frac{1}{61}< \frac{1}{60}\)

                \(\frac{1}{84}< \frac{1}{60}\)

                \(\frac{1}{96}< \frac{1}{60}\)

\(=>\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}\)

\(=>\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{60}.3\left(2\right)\)

Từ (1) và (2) suy ra: \(\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)

\(=>\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{5}+3.\left(\frac{1}{12}+\frac{1}{60}\right)\)

\(=>\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{2}\)

\(=>Đpcm\)