Xét $n=2$:
$\dfrac{1}{2+1}+\dfrac{1}{2+2}>\dfrac{13}{24}$ (đúng)
Giả sử bất đẳng thức đúng với $n=k$ ($k\ge 2$):
$\dfrac{1}{k+1}+\dfrac{1}{k+2}+...+\dfrac{1}{k+k}>\dfrac{13}{24}$
CMR bất đẳng thức đúng với $n=k+1$:
$\dfrac{1}{k+1+1}+\dfrac{1}{k+1+2}+...+\dfrac{1}{k+1+k+1}>\dfrac{13}{24}$
$\Leftrightarrow \dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{k+k+2}>\dfrac{13}{24}$
Thật vậy:
$VT=\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{k+k}+\dfrac{1}{k+k+1}+\dfrac{1}{k+k+2}$
$>\dfrac{13}{24}-\dfrac{1}{k+1}+\dfrac{1}{k+k+1}+\dfrac{1}{k+k+2}$
$=\dfrac{13}{24}-\dfrac{1}{k+1}+\dfrac{1}{2k+1}+\dfrac{0,5}{k+1}$
$=\dfrac{13}{24}-\dfrac{1}{2k+2}+\dfrac{1}{2k+1}$
$=\dfrac{13}{24}+\dfrac{-2k-1+2k+2}{(2k+2)(2k+1)}$
$=\dfrac{13}{24}+\dfrac{1}{(2k+2)(2k+1)}>\dfrac{13}{24}$
$>VP$
$\to$ đpcm
