Geometrical method

**LuyenTDH** · April 3rd 2011, 03:48 AM

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.

**veileen** · April 3rd 2011, 04:18 AM

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.

**LuyenTDH** · April 3rd 2011, 04:33 AM

Thanks for Veileen's soon comments.

By chance, do you know about the Titu's lemma?

**veileen** · April 3rd 2011, 04:39 AM

"Titu's lemma" Do you refer at that lemma that can be deduced from Cauchy's inequality? If so, yes, I know ^^

**LuyenTDH** · April 3rd 2011, 04:51 AM

Originally Posted by veileen

"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?

**veileen** · April 3rd 2011, 05:06 AM

$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^ 2}{b_n}\geq \frac{(a_1+a_2+...+a_n)^2}{b_1+b_+...+b_n},$
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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