cm bdt schwarz

V

vy000

Tổng quát của schwarz:

$\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+...+\dfrac{a_n^2}{b_n} \ge \dfrac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}$ với $b_1;b_2;...;b_n>0$

Xét BĐT bunhiacopsky:

$(\dfrac{a_1}{\sqrt{b_1}}\sqrt{b_1}+\dfrac{a_2}{ \sqrt{b_2}}\sqrt{b_2}+...\dfrac{a_n}{\sqrt{b_n}} \sqrt{b_n})^2 \le (\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+...+\dfrac{a_n^2}{b_n})(b_1+b_2+...+b_n)$

Chứng tỏ ...
 
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