已知 $a, b > 0$，求证：$\dfrac{1}{{a + b}} + \dfrac{1}{{a + 2b}} + \cdots + \dfrac{1}{{a + nb}} < \dfrac{n}{{\sqrt {\left( {a + \frac{1}{2}b} \right)\left( {a + \frac{{2n + 1}}{2}b} \right)} }}$．
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【答案】
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【解析】
记不等式左侧为 $M$，则\[\begin{split}M^2&=\left( {\dfrac{1}{{a + b}} + \dfrac{1}{{a + 2b}} + \cdots + \dfrac{1}{{a + nb}}} \right)^2\\&\leqslant n \cdot \left[ {\dfrac{1}{{{{\left( {a + b} \right)}^2}}} + \dfrac{1}{{{{\left( {a + 2b} \right)}^2}}} + \cdots + \dfrac{1}{{{{\left( {a + nb} \right)}^2}}}} \right]\\&< n \cdot \left[ {\dfrac{1}{{\left( {a + \frac{1}{2}b} \right)\left( {a + \frac{3}{2}b} \right)}} + \dfrac{1}{{\left( {a + \frac{3}{2}b} \right)\left( {a + \frac{5}{2}b} \right)}} + \cdots \dfrac{1}{{\left( {a + \frac{{2n - 1}}{2}b} \right)\left( {a + \frac{{2n + 1}}{2}b} \right)}}} \right]\\&= \dfrac{n}{b} \cdot \left( {\dfrac{1}{{a + \frac{1}{2}b}} - \dfrac{1}{{a + \frac{{2n + 1}}{2}b}}} \right)\\&= \dfrac{n}{b} \cdot \dfrac{{nb}}{{\left( {a + \frac{1}{2}b} \right)\left( {a + \frac{{2n + 1}}{2}b} \right)}} \\&= \dfrac{{{n^2}}}{{\left( {a + \frac{1}{2}b} \right)\left( {a + \frac{{2n + 1}}{2}b} \right)}},\end{split}\]所以原命题得证．
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