Chủ đề này có 3 trả lời

[DOTOANNANG]

$\lceil$ https://diendantoanh...dt/#entry719948 $\rfloor$

$\lceil$ To*Arthur*Pendragon $\it{!}$    $\rfloor$

$\sum\,\frac{\it{3}}{\it{5}- \it{6}\,\it{ab}}=$ $\sum\,\frac{\it{3}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}}{\it{5}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}- \it{6}\,\it{ab}}\leqq$ $\sum\,\frac{\it{6}\,\it{c}^{\,\it{2}}+ \it{12}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}\,\,\it{)}+ \it{(}\,\,\sqrt{\it{629}}- \it{5}\,\,\it{)}\it{(}\,\,\it{bc}+ \it{ca}\,\,\it{)}+ \it{(}\,\,\it{91}- \it{2}\sqrt{\it{629}}\,\,\it{)}\it{ab}}{\it{27}\it{(}\,\,\it{ab}+ \it{bc}+ \it{ca}\,\,\it{)}+ \it{10}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}}=$ $\it{3}$

Bài viết đã được chỉnh sửa nội dung bởi DOTOANNANG: 07-02-2019 - 14:58

[DOTOANNANG]

$\lceil$ https://diendantoanh...dt/#entry719948 $\rfloor$

$\lceil$ To*Arthur*Pendragon $\it{!}$    $\rfloor$

$\sum\,\frac{\it{3}}{\it{5}- \it{6}\,\it{ab}}=$ $\sum\,\frac{\it{3}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}}{\it{5}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}- \it{6}\,\it{ab}}\leqq$ $\sum\,\frac{\it{6}\,\it{c}^{\,\it{2}}+ \it{12}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}\,\,\it{)}+ \it{(}\,\,\sqrt{\it{629}}- \it{5}\,\,\it{)}\it{(}\,\,\it{bc}+ \it{ca}\,\,\it{)}+ \it{(}\,\,\it{91}- \it{2}\sqrt{\it{629}}\,\,\it{)}\it{ab}}{\it{27}\it{(}\,\,\it{ab}+ \it{bc}+ \it{ca}\,\,\it{)}+ \it{10}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}}=$ $\it{3}$

$\sum\,\frac{\it{3}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}}{\it{5}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}- \it{6}\,\it{ab}}\leqq$ $\sum\,\frac{\it{18}\,\it{c}^{\,\it{2}}+ \it{6}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}\,\,\it{)}+ \it{(}\,\,\sqrt{\it{341}}- \it{5}\,\,\it{)}\it{(}\,\,\it{bc}+ \it{ca}\,\,\it{)}+ \it{(}\,\,\it{73}- \it{2}\sqrt{\it{341}}\,\,\it{)}\it{ab}}{\it{21}\it{(}\,\,\it{ab}+ \it{bc}+ \it{ca}\,\,\it{)}+ \it{10}\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}\,\,\it{)}}=$ $\it{3}$

Áp dụng $\it{ab}\leqq \left | \it{a} \right |\,.\,\left | \it{b} \right |$ cho bài trên nên chỉ xét với $\it{a},\,\it{b},\,\it{c}\geqq \it{0}$ $\it{!}$ 

Bài viết đã được chỉnh sửa nội dung bởi DOTOANNANG: 07-02-2019 - 16:21

[DOTOANNANG]

Nhận xét ngay $\it{:}$

$$\sum\,\frac{\it{1}}{\it{5}- \it{6}\,\it{bc}}\leqq \sum\,\left ( \frac{\it{1}}{\it{3}}- \frac{\it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}}{\it{3}}+ \frac{\it{b}^{\,\it{4}}+ \it{c}^{\,\it{4}}}{\it{3}}+ \frac{\it{4}\,\it{b}^{\,\it{2}}\it{c}^{\,\it{2}}}{\it{3}} \right )= \it{1}$$

[DOTOANNANG]

Xem thêm về phân tích  $\lceil$ Schur*decomposition $\rfloor$ cho bất đẳng thức trên tại $\it{:}$

$\lceil$ https://h-a-i-d-a-n-...19/02/09/135020 $\rfloor$