Ta có
$VT = \dfrac{1}{3(1 + \sqrt{2})} + \dfrac{1}{5(\sqrt{2} + \sqrt{3})} + \cdots + \dfrac{1}{4027(\sqrt{2013} + \sqrt{2014})}$
$= \dfrac{\sqrt{2} - 1}{3} + \dfrac{\sqrt{3} - \sqrt{2}}{5} + \dfrac{\sqrt{4} - \sqrt{3}}{7} + \cdots + \dfrac{\sqrt{2014} - \sqrt{2013}}{4027}$
$= \dfrac{\sqrt{2} - \sqrt{1}}{2 + 1} + \dfrac{\sqrt{3} - \sqrt{2}}{3 + 2} + \dfrac{\sqrt{4} + \sqrt{3}}{4 + 3} + \cdots + \dfrac{\sqrt{2014} - \sqrt{2013}}{2014 + 2013}$
Ta thấy các số hạng ở vế trái đều có dạng $\dfrac{\sqrt{a} - \sqrt{b}}{a + b}$
Áp dụng BĐT Cauchy ta có
$a + b \geq 2\sqrt{ab}$
$<-> \dfrac{\sqrt{a} - \sqrt{b}}{a + b} \leq \dfrac{\sqrt{a} - \sqrt{b}}{2\sqrt{ab}} = \dfrac{1}{2} \left( \dfrac{1}{\sqrt{b}} - \dfrac{1}{\sqrt{a}} \right)$
Do đó
$2.VT \leq 1 - \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{3}} + \dfrac{1}{\sqrt{3}} - \dfrac{1}{\sqrt{4}} + \cdots + \dfrac{1}{\sqrt{2013}} - \dfrac{1}{\sqrt{2014}}$
$<-> 2 VT \leq 1 - \dfrac{1}{\sqrt{2014}}$
$<-> VT \leq \dfrac{1}{2} - \dfrac{1}{2\sqrt{2014}} < \dfrac{1}{2} < \dfrac{2013}{2014}$
Ta có đpcm.
