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]}