Ta có $a^2>a^2-1=a^2-ab+ab-1=(a-1)(a+1)$
Từ đó ta có $\dfrac{1}{a^2}<\dfrac{1}{(a-1)(a+1)}$
$\begin{array}{l}
A = 1 + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + ... + \dfrac{1}{{{{2016}^2}}}\\
= 1 + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + ... + \dfrac{1}{{{{2016}^2}}} < B = 1 + \dfrac{1}{{1.3}} + \dfrac{1}{{2.4}} + \dfrac{1}{{3.5}} + ... + \dfrac{1}{{2014.2016}}\\
B = 1 + \dfrac{1}{2}.\left( {\dfrac{1}{1} - \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{3} - \dfrac{1}{5} + ... + \dfrac{1}{{2014}} - \dfrac{1}{{2016}}} \right) = 1 + \dfrac{1}{2}\left( {1 + \dfrac{1}{2} - \dfrac{1}{{2015}} - \dfrac{1}{{2017}}} \right) < 1 + \dfrac{1}{2}.\dfrac{3}{2} = \dfrac{7}{4}
\end{array}$
