Cho $a_1;a_2;...;a_n >0$. Thoả $a_1+a_2+...+a_n =1$
$$\left({\frac{1}{{a_1^2}}-1}\right)\left({\frac{1}{{a_2^2}}-1}\right)...\left({\frac{1}{{a_n^2}}-1}\right)\geqslant{({n^2}-1)^n}$$
Cho $a_1;a_2;...;a_n >0$. Thoả $a_1+a_2+...+a_n =1$
$$\left({\frac{1}{{a_1^2}}-1}\right)\left({\frac{1}{{a_2^2}}-1}\right)...\left({\frac{1}{{a_n^2}}-1}\right)\geqslant{({n^2}-1)^n}$$
Cho $a_1;a_2;...;a_n >0$. Thoả $a_1+a_2+...+a_n =1$
$$\left({\frac{1}{{a_1^2}}-1}\right)\left({\frac{1}{{a_2^2}}-1}\right)...\left({\frac{1}{{a_n^2}}-1}\right)\geqslant{({n^2}-1)^n}$$
bđt$<=> \prod (1-a_{1}).\prod (1+a_{1}) \geq (n^{2}-1)^{n}.\prod a_{1}^{2}$
$<=> \prod (a_{2}+a_{3}+...+a_n) .\prod (a_1+a_2+...+a_n+a_1) \geq (n^{2}-1)^{2}.\prod a_1^{2}$
(đúng theo cô-si)
Bài viết đã được chỉnh sửa nội dung bởi canhhoang30011999: 16-08-2014 - 16:57
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