Lema Titu (ditemukan oleh Titu Andreescu , atau dikenal juga lema T2 , bentuk Engel, atau pertidaksamaan Sedrakyan ) menyatakan untuk real positif, kita harus mencari
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{\displaystyle {\frac {\left(\sum _{i=1}^{n}u_{i}\right)^{2}}{\sum _{i=1}^{n}v_{i}}}\leq \sum _{i=1}^{n}{\frac {u_{i}^{2}}{v_{i}}}.}
Konsekuensi dari Pertidaksamaan Cauchy-Schwarz adalah perolehan setelah menggunakan
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{\displaystyle u_{i}'={\frac {u_{i}}{\sqrt {v_{i}}}}}
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{\displaystyle v_{i}'={\sqrt {v_{i}}}.}
Maka
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{\displaystyle {\begin{aligned}\left({\frac {x_{1}^{2}}{y_{1}}}+{\frac {x_{2}^{2}}{y_{2}}}+\cdots +{\frac {x_{n}^{2}}{y_{n}}}\right)(y_{1}+y_{2}+\cdots +y_{n})&\geq (x_{1}+x_{2}+\cdots +x_{n})^{2}\\{\frac {x_{1}^{2}}{y_{1}}}+{\frac {x_{2}^{2}}{y_{2}}}+\cdots +{\frac {x_{n}^{2}}{y_{n}}}&\geq {\frac {(x_{1}+x_{2}+\cdots +x_{n})^{2}}{(y_{1}+y_{2}+\cdots +y_{n})}}.\end{aligned}}}
Dari nilai biasa
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{\displaystyle a_{1},a_{2},\ldots ,a_{n}}
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{\displaystyle b_{1},b_{2},\ldots ,b_{n}}
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{\displaystyle {\frac {a_{1}^{2}}{b_{1}}}+{\frac {a_{2}^{2}}{b_{2}}}+\cdots +{\frac {a_{n}^{2}}{b_{n}}}\geq {\frac {(a_{1}+a_{2}+\cdots +a_{n})^{2}}{b_{1}+b_{2}+\cdots +b_{n}}}.}
Setelah itu memperoleh dengan menerapkan substitusi
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{\displaystyle a_{i}={\frac {x_{i}}{\sqrt {y_{i}}}}}
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{\displaystyle b_{i}={\sqrt {y_{i}}}}
pertidaksamaan Cauchy-Schwarz .
Maka
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{\displaystyle {\begin{aligned}\left({\frac {x_{1}^{2}}{y_{1}}}+{\frac {x_{2}^{2}}{y_{2}}}+\cdots +{\frac {x_{n}^{2}}{y_{n}}}\right)(y_{1}+y_{2}+\cdots +y_{n})&\geq (x_{1}+x_{2}+\cdots +x_{n})^{2}\\{\frac {x_{1}^{2}}{y_{1}}}+{\frac {x_{2}^{2}}{y_{2}}}+\cdots +{\frac {x_{n}^{2}}{y_{n}}}&\geq {\frac {(x_{1}+x_{2}+\cdots +x_{n})^{2}}{(y_{1}+y_{2}+\cdots +y_{n})}}.\ _{\square }\end{aligned}}}
Lemma Titu, konsekuensi langsung dari pertidaksamaan Cauchy–Schwarz , menyatakan bahwa untuk setiap urutan
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{\displaystyle n}
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{\displaystyle (x_{k})}
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{\displaystyle (a_{k})}
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{\displaystyle \displaystyle \sum _{k=1}^{n}{\frac {x_{k}^{2}}{a_{k}}}\geq {\frac {(\sum _{k=1}^{n}x_{k})^{2}}{\sum _{k=1}^{n}a_{k}}}}
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{\displaystyle a,b,c}
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{\displaystyle a(b+c),b(c+a),c(a+b)}
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{\displaystyle {\frac {a^{2}}{a(b+c)}}+{\frac {b^{2}}{b(c+a)}}+{\frac {c^{2}}{c(a+b)}}\geq {\frac {(a+b+c)^{2}}{a(b+c)+b(c+a)+c(a+b)}}}
Dengan mengalikan semua hasil kali di sisi yang lebih kecil dan mengumpulkan suku-suku sejenis, kita memperoleh
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{\displaystyle {\frac {a^{2}}{a(b+c)}}+{\frac {b^{2}}{b(c+a)}}+{\frac {c^{2}}{c(a+b)}}\geq {\frac {a^{2}+b^{2}+c^{2}+2(ab+bc+ca)}{2(ab+bc+ca)}},}
yang disederhanakan menjadi
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{\displaystyle {\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}\geq {\frac {a^{2}+b^{2}+c^{2}}{2(ab+bc+ca)}}+1.}
Dengan pertidaksamaan penataan ulang , maka
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{\displaystyle a^{2}+b^{2}+c^{2}\geq ab+bc+ca}
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{\displaystyle \displaystyle {\frac {1}{2}}}
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{\displaystyle {\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}
Jika nilai
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{\displaystyle \displaystyle {\bigg (}\sum _{i=1}^{n}{\sqrt[{\frac {m}{m-1}}]{a_{i}}}^{\frac {m}{m-1}}{\bigg )}^{\frac {1}{\ {\frac {m}{m-1}}\ }}\left(\sum _{i=1}^{n}{\dfrac {x_{i}^{m}}{\left({\frac {a_{i}}{\sqrt[{m}]{a_{i}}}}\right)^{m}}}\right)^{\frac {1}{m}}\geq \sum _{i=1}^{n}x_{i},}
pertidaksamaan Holder
Maka menyederhanakan hasil, yaitu:
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{\displaystyle {\begin{aligned}\displaystyle {\bigg (}\sum _{i=1}^{n}a_{i}{\bigg )}^{\frac {m-1}{m}}{\bigg (}\sum _{i=1}^{n}{\dfrac {x_{i}^{m}}{a_{i}^{m-1}}}{\bigg )}^{\frac {1}{m}}&\geq \sum _{i=1}^{n}x_{i}\\{\bigg (}\sum _{i=1}^{n}a_{i}{\bigg )}^{m-1}{\bigg (}\sum _{i=1}^{n}{\dfrac {x_{i}^{m}}{a_{i}^{m-1}}}{\bigg )}&\geq {\bigg (}\sum _{i=1}^{n}x_{i}{\bigg )}^{m}\\{\bigg (}\sum _{i=1}^{n}{\dfrac {x_{i}^{m}}{a_{i}^{m-1}}}{\bigg )}&\geq {\dfrac {{\bigg (}\sum _{i=1}^{n}x_{i}{\bigg )}^{m}}{{\bigg (}\sum _{i=1}^{n}a_{i}{\bigg )}^{m-1}}}.\end{aligned}}}
![{\displaystyle \displaystyle {\bigg (}\sum _{i=1}^{n}{\sqrt[{\frac {m}{m-1}}]{a_{i}}}^{\frac {m}{m-1}}{\bigg )}^{\frac {1}{\ {\frac {m}{m-1}}\ }}\left(\sum _{i=1}^{n}{\dfrac {x_{i}^{m}}{\left({\frac {a_{i}}{\sqrt[{m}]{a_{i}}}}\right)^{m}}}\right)^{\frac {1}{m}}\geq \sum _{i=1}^{n}x_{i},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37b9315818a2f25da041c0e36c086e7294496e89)
