[Toán 10].Bất Đẳng Thức

N

noinhobinhyen

BĐT Cô-Si :

$a_1+a_2+a_3+...+a_n \geq n\sqrt[n]{a_1a_2a_3...a_n}$

$\forall a_1;a_2;...;a_n \geq 0$

BĐT Bu-nhi-a-cop-xki :

$(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2) \geq (a_1b_1+a_2b_2+...+a_nb_n)^2$
 
You must log in or register to reply here.
Top Bottom