Với $n=2$: $\left.\begin{align}&\text{VT}=\dfrac{1}{1^2}+\dfrac{1}{2^2}=\dfrac{5}{4}\\ &\text{VP}=2-\dfrac{1}{2}=\dfrac{3}{2}\end{align}\right\}\Rightarrow$ BĐT đúng

Giả sử BĐT đúng với $n=k\ge2$, tức là $$\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{k^2}<2-\dfrac{1}{k}$$
Ta sẽ chứng minh BĐT cũng đúng với $n=k+1$, tức là
$$\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{k^2}+\dfrac{1}{(k+1)^2}<2-\dfrac{1}{k+1}$$
Thật vậy:
$\Leftrightarrow\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{k^2}<2-\dfrac{1}{k}$ (theo giả thiết quy nạp)
$\Leftrightarrow\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{k^2}+\dfrac{1}{(k+1)^2}<\left(2-\dfrac{1}{k} \right)+\dfrac{1}{(k+1)^2}$
$\Leftrightarrow\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{k^2}+\dfrac{1}{(k+1)^2}<2-\left(\dfrac{1}{k}-\dfrac{1}{(k+1)^2}\right)$
$\Leftrightarrow\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{k^2}+\dfrac{1}{(k+1)^2}<2-\dfrac{(k+1)^2-k}{k(k+1)^2}$
$\Leftrightarrow\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{k^2}+\dfrac{1}{(k+1)^2}<2-\dfrac{k^2+k+1}{k(k+1)^2}$
$\Leftrightarrow\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{k^2}+\dfrac{1}{(k+1)^2}<2-\dfrac{k^2+k}{k(k+1)^2}$
$\Leftrightarrow\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{k^2}+\dfrac{1}{(k+1)^2}<2-\dfrac{k(k+1)}{k(k+1)^2}$
$\Leftrightarrow\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{k^2}+\dfrac{1}{(k+1)^2}<2-\dfrac{1}{k+1}$
$\Rightarrow$ BĐT đúng với $n=k+1$
Vậy BĐT đúng với $\forall n\in\mathbb{N}^*$ và $n\ge2$.