放縮法是通過捨去或添加一些項來構造不等式的一種方法。[參 1] 本條目存在以下問題,請協助改善本條目或在討論頁針對議題發表看法。 此條目包含過多行話或專業術語,可能需要簡化或提出進一步解釋。 (2014年7月16日) 此條目需要編修,以確保文法、用詞、語氣、格式、標點等使用恰當。 (2014年7月16日) 例子 求證 log 2 3 + log 3 2 < 2 + 1 {\displaystyle {\sqrt {\log _{2}3}}+{\sqrt {\log _{3}2}}<{\sqrt {2}}+1} [注 1] log 2 3 + log 3 2 < log 2 4 + log 3 3 = 2 + 1 {\displaystyle {\sqrt {\log _{2}3}}+{\sqrt {\log _{3}2}}<{\sqrt {\log _{2}4}}+{\sqrt {\log _{3}3}}={\sqrt {2}}+1} [參 2] 已知a,b,c,d為正數,求證 1 < a a + b + d + b a + b + c + c b + c + d + d a + c + d < 2 {\displaystyle 1<{\frac {a}{a+b+d}}+{\frac {b}{a+b+c}}+{\frac {c}{b+c+d}}+{\frac {d}{a+c+d}}<2} a a + b + d + b a + b + c + c b + c + d + d a + c + d > a a + b + c + d + b a + b + c + d + c a + b + c + d + 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+c+d}}+{\frac {b}{a+b+c+d}}+{\frac {c}{a+b+c+d}}+{\frac {d}{a+b+c+d}}=1} a a + b + d + b a + b + c + c b + c + d + d a + c + d < a a + b + b a + b + c c + d + d 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}}+{\frac {b}{a+b}}+{\frac {c}{c+d}}+{\frac {d}{c+d}}=2} [參 3] 求證 ∑ k = 1 n 1 k 2 < 2 − 1 n {\displaystyle \sum _{k=1}^{n}{\frac {1}{k^{2}}}<2-{\frac {1}{n}}} ∑ k = 1 n 1 k 2 < 1 + ∑ k = 2 n 1 k ( k − 1 ) = 1 + ∑ k = 2 n 1 k − 1 − 1 k = 2 − 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] [參 2] 設 n ∈ N + {\displaystyle n\in N^{+}} ,求證 n ( n + 1 ) 2 < ∑ k = 1 n k ( k + 1 ) < ( n + 1 ) 2 2 {\displaystyle {\frac {n(n+1)}{2}}<\sum _{k=1}^{n}{\sqrt {k(k+1)}}<{\frac {(n+1)^{2}}{2}}} k = k 2 < k ( k + 1 ) < k 2 + k + 1 4 = k + 1 2 {\displaystyle k={\sqrt {k^{2}}}<{\sqrt {k(k+1)}}<{\sqrt {k^{2}+k+{\frac {1}{4}}}}=k+{\frac {1}{2}}} n ( n + 1 ) 2 = ∑ k = 1 n k < ∑ k = 1 n k ( k + 1 ) < ∑ k = 1 n ( k + 1 2 ) = n 2 + 2 n 2 < ( n + 1 ) 2 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] [參 3] 設a,b,c為直角三角形的三邊,c為斜邊,求證: a n + b n < c n ( n > 2 ) {\displaystyle a^{n}+b^{n}<c^{n}(n>2)} 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 ) {\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] [參 2] Remove ads備註 [注 1]log為對數函數 [注 2]這裡用了裂項和的求和方法 [注 3]這裡用了等冪求和的求和方法 [注 4]這裡用了勾股定理 參考資料 [參 1]用放缩法证明不等式. (原始內容存檔於2014-07-26). [參 2]董衛平. 说说放缩法. 數學大世界(高中). 2011, (2) [2015-09-20]. (原始內容存檔於2016-03-04). [參 3]例谈不等式证明的十种常用方法. [2014-07-16]. (原始內容存檔於2014-07-20). Loading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.Remove ads