放缩法

放缩法是通过舍去或添加一些项来构造不等式的一种方法。^{[参 1]}

例子

1. 求证${\displaystyle {\sqrt {\log _{2}3}}+{\sqrt {\log _{3}2}}<{\sqrt {2}}+1}$ ^{[注 1]}

   ${\displaystyle {\sqrt {\log _{2}3}}+{\sqrt {\log _{3}2}}<{\sqrt {\log _{2}4}}+{\sqrt {\log _{3}3}}={\sqrt {2}}+1}$ ^{[参 2]}

2. 已知a,b,c,d为正数，求证${\displaystyle 1<{\frac {a}{a+b+d}}+{\frac {b}{a+b+c}}+{\frac {c}{b+c+d}}+{\frac {d}{a+c+d}}<2}$ 

   ${\displaystyle {\frac {a}{a+b+d}}+{\frac {b}{a+b+c}}+{\frac {c}{b+c+d}}+{\frac {d}{a+c+d}}>{\frac {a}{a+b+c+d}}+{\frac {b}{a+b+c+d}}+{\frac {c}{a+b+c+d}}+{\frac {d}{a+b+c+d}}=1}$ 

   ${\displaystyle {\frac {a}{a+b+d}}+{\frac {b}{a+b+c}}+{\frac {c}{b+c+d}}+{\frac {d}{a+c+d}}<{\frac {a}{a+b}}+{\frac {b}{a+b}}+{\frac {c}{c+d}}+{\frac {d}{c+d}}=2}$ ^{[参 3]}

3. 求证${\displaystyle \sum _{k=1}^{n}{\frac {1}{k^{2}}}<2-{\frac {1}{n}}}$ 

   ${\displaystyle \sum _{k=1}^{n}{\frac {1}{k^{2}}}<1+\sum _{k=2}^{n}{\frac {1}{k(k-1)}}=1+\sum _{k=2}^{n}{\frac {1}{k-1}}-{\frac {1}{k}}=2-{\frac {1}{n}}}$ ^{[注 2] [参 2]}

4. 设${\displaystyle n\in N^{+}}$ ，求证${\displaystyle {\frac {n(n+1)}{2}}<\sum _{k=1}^{n}{\sqrt {k(k+1)}}<{\frac {(n+1)^{2}}{2}}}$ 

   ${\displaystyle k={\sqrt {k^{2}}}<{\sqrt {k(k+1)}}<{\sqrt {k^{2}+k+{\frac {1}{4}}}}=k+{\frac {1}{2}}}$ 

   ${\displaystyle {\frac {n(n+1)}{2}}=\sum _{k=1}^{n}k<\sum _{k=1}^{n}{\sqrt {k(k+1)}}<\sum _{k=1}^{n}(k+{\frac {1}{2}})={\frac {n^{2}+2n}{2}}<{\frac {(n+1)^{2}}{2}}}$ ^{[注 3] [参 3]}

5. 设a,b,c为直角三角形的三边,c为斜边，求证：${\displaystyle a^{n}+b^{n}<c^{n}(n>2)}$ 

   ${\displaystyle a^{n}+b^{n}=a^{2}a^{n-2}+b^{2}b^{n-2}<a^{2}c^{n-2}+b^{2}c^{n-2}=c^{n}(n>2)}$ ^{[注 4] [参 2]}

备注

1. log为对数函数
2. 这里用了裂项和的求和方法
3. 这里用了等幂求和的求和方法
4. 这里用了勾股定理

参考资料

1. . （原始内容存档于2014-07-26）. 
2. 董卫平. 说说放缩法. 数学大世界(高中). 2011, (2) [2015-09-20]. （原始内容于2016-03-04）. 
3. 例谈不等式证明的十种常用方法. [2014-07-16]. （原始内容于2014-07-20）.