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Math Help - Geometrical method

  1. #1
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    Post Geometrical method

    I was said that a geometrical method is very useful to solve problems. Do you have any idea? Any book can you suggest me to read?

    Thanks.
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  2. #2
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    Well, if "geometrical" includes trigonometry and vectors I have some problems that can be solved easier like that. Also geometrical method is useful in problems with complex numbers (and, yes, I have problems that can be solved easier that way too).

    Books - um, a good and complet one that I know is "Complex numbers from A to Z" - Titu Andreescu, Dorin Andrica.
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  3. #3
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    Post Geometrical methods

    Thanks for Veileen's soon comments.

    By chance, do you know about the Titu's lemma?
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  4. #4
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    "Titu's lemma" Do you refer at that lemma that can be deduced from Cauchy's inequality? If so, yes, I know ^^
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  5. #5
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    Titu's lemma

    Quote Originally Posted by veileen View Post
    "Titu's lemma" Do you refer at that lemma that can be deduced from Cauchy's inequality? If so, yes, I know ^^
    Yes. Could you show me?
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  6. #6
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    \frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^  2}{b_n}\geq \frac{(a_1+a_2+...+a_n)^2}{b_1+b_+...+b_n},
    Equality if and only if \frac{a_1}{b_1}=\frac{a_2}{b_2}=...=\frac{a_n}{b_n  }

    Demonstration:
    (x_1^2+x_2^2+...+x_n^2)(y_1^2+y_2^2+...+y_n^2)\geq (x_1y_1+x_2y_2+...+x_ny_n)^2 (Cauchy's inequality)

    Let x_i=\frac{a_i}{\sqrt{b_i}} and y_i=\sqrt{b_i}, then:

    \sum_{i=1}^{n}\frac{a_i^2}{{b_i}}\cdot \sum_{i=1}^{n}b_i\geq \left ( \sum_{i=1}^{n}a_i^2 \right )\Rightarrow \sum_{i=1}^{n}\frac{a_i^2}{{b_i}}\geq \frac{\left ( \sum_{i=1}^{n}a_i^2 \right )}{\sum_{i=1}^{n}b_i} which is exactly Titu's inequality.
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