Với $n=2$: $\left.\begin{align}& VT=\dfrac{1}{2+1}+\dfrac{1}{2+2}=\dfrac{7}{12} \\ & VP=\dfrac{13}{24}\end{align}\right\}\Rightarrow$ BĐT đúng

Giả sử BĐT đúng với $n=k\ge 2$, tức là $$\dfrac{1}{k+1}+\dfrac{1}{k+2}+\cdots+\dfrac{1}{k+k}>\dfrac{13}{24}$$
Ta sẽ chứng minh BĐT cũng đúng với $n=k+1$, tức là
$$\dfrac{1}{k+2}+\dfrac{1}{k+3}+\cdots+\dfrac{1}{k+k}+\dfrac{1}{k+k+1}+\dfrac{1}{(k+1)+(k+1)}>\dfrac{13}{24}$$
Thật vậy, theo giả thiết quy nạp ta có
$\Leftrightarrow\dfrac{1}{k+1}+\dfrac{1}{k+2}+\cdots+\dfrac{1}{k+k}>\dfrac{13}{24}$
$\Leftrightarrow\dfrac{1}{k+1}+\dfrac{1}{k+2}+\cdots+\dfrac{1}{k+k}+\left(\dfrac{1}{k+k+1}+\dfrac{1}{(k+1)+(k+1)}-\dfrac{1}{k+1}\right)>\dfrac{13}{24}+\left( \dfrac{1}{k+k+1}+\dfrac{1}{(k+1)+(k+1)}-\dfrac{1}{k+1} \right)$
$\Leftrightarrow\dfrac{1}{k+2}+\dfrac{1}{k+3}+\cdots+\dfrac{1}{k+k}+\dfrac{1}{k+k+1}+\dfrac{1}{(k+1)+(k+1)}>\dfrac{13}{24}+\left(\dfrac{1}{2k+1}+\dfrac{1}{2(k+1)}-\dfrac{1}{k+1} \right)$
$\Leftrightarrow\dfrac{1}{k+2}+\dfrac{1}{k+3}+\cdots+\dfrac{1}{k+k}+\dfrac{1}{k+k+1}+\dfrac{1}{(k+1)+(k+1)}>\dfrac{13}{24}+\dfrac{2(k+1)+2k+1-2(2k+1)}{2(k+1)(2k+1)}$
$\Leftrightarrow\dfrac{1}{k+2}+\dfrac{1}{k+3}+\cdots+\dfrac{1}{k+k}+\dfrac{1}{k+k+1}+\dfrac{1}{(k+1)+(k+1)}>\dfrac{13}{24}+\dfrac{1}{2(k+1)(2k+1)}$
$\Leftrightarrow\dfrac{1}{k+2}+\dfrac{1}{k+3}+\cdots+\dfrac{1}{k+k}+\dfrac{1}{k+k+1}+\dfrac{1}{(k+1)+(k+1)}>\dfrac{13}{24}$
$\Rightarrow $ BĐT đúng với $n=k+1$

Vậy BĐT đúng với $\forall n\in\mathbb{N}^*$ và $n\ge2$.