Sử dụng$[$hệ thức lớp $\it{9}$$]$$:$ $\sqrt{\it{A}}- \it{B}= \frac{\it{A}- \it{B}^{\,\it{2}}}{\sqrt{\it{A}}+ \it{B}}$$,$ có thể tái bất đẳng thức một số$:$ $\lceil$ https://diendantoanh...e-1#entry721183 $\rfloor$
Sử dụng $\sqrt{\it{A}}- \it{B}= \frac{\it{A}- \it{B}^{\,\it{2}}}{\sqrt{\it{A}}+ \it{B}}$$:$
$- \it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{bc}+ \it{ca}\,\,\it{)}+ \it{2}\sqrt{\it{(}\,\,\it{a}^{\,\it{2}}+ \it{bc}\,\,\it{)}\it{(}\,\,\it{b}^{\,\it{2}}+ \it{ca}\,\,\it{)}}=$
$= \frac{-\it{(}\,\,\it{a}- \it{b}\,\,\it{)}^{\,\it{2}}\it{(}\,\,\it{a}+ \it{b}- \it{c}\,\,\it{)}^{\,\it{2}}}{\it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{bc}+ \it{ca}\,\,\it{)}+ \it{2}\sqrt{\it{(}\,\,\it{a}^{\,\it{2}}+ \it{bc}\,\,\it{)}\it{(}\,\,\it{b}^{\,\it{2}}+ \it{ca}\,\,\it{)}}}\leqq$
$\leqq \frac{-\it{(}\,\,\it{a}- \it{b}\,\,\it{)}^{\,\it{2}}\it{(}\,\,\it{a}+ \it{b}- \it{c}\,\,\it{)}^{\,\it{2}}}{\it{2}\it{(}\,\,\it{a}+ \it{b}+ \it{c}\,\,\it{)}^{\,\it{2}}}$
Kết hợp$:$
$\sum\,\sqrt{\it{a}^{\,\it{2}}+ \it{bc}}\leqq \frac{\it{3}}{\it{2}}\sum\,\it{a}\Leftrightarrow \sum\,\it{a}^{\,\it{2}}+ \sum\,\it{bc}+ \it{2}\,\sum\,\sqrt{\it{(}\,\,\it{a}^{\,\it{2}}+ \it{bc}\,\,\it{)}\it{(}\,\,\it{b}^{\,\it{2}}+ \it{ca}\,\,\it{)}}\leqq \frac{\it{9}}{\it{4}}\it{(}\,\,\sum\,\it{a}\,\,\it{)}^{\,\it{2}}$$.$ Và ta cần phải chứng minh$:$
$\sum\,\it{a}^{\,\it{2}}+ \sum\,\it{bc}+ \sum\left ( \it{(}\,\,\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{bc}+ \it{ca}\,\,\it{)}+ \frac{-\it{(}\,\,\it{a}- \it{b}\,\,\it{)}^{\,\it{2}}\it{(}\,\,\it{a}+ \it{b}- \it{c}\,\,\it{)}^{\,\it{2}}}{\it{2}\it{(}\,\,\it{a}+ \it{b}+ \it{c}\,\,\it{)}^{\,\it{2}}} \right )\leqq \frac{\it{9}}{\it{4}}\it{(}\,\,\sum\,\it{a}\,\,\it{)}^{\,\it{2}}$
Bất đẳng thức đối xứng thuần bậc $\it{4}$$:$ $\lceil$ https://ajmaa.org/se...9n1/v9i1p15.pdf $\rfloor$ hoặc có thể giải bằng B$\ast$W và S$\ast$O$\ast$S$.$
Hiện tại mình đang nghiên cứu để có thể viết$($và chưa được$)$$:$
$$\sqrt{\it{a}^{\,\it{2}}+ \it{bc}}\leqq \frac{\it{3}\,\it{F}\it{(}\,\,\it{a},\,\it{b},\,\it{c}\,\,\it{)}\sum\,\it{a}}{\it{2}\,\sum\limits_{cyc}\it{F}\it{(}\,\,\it{a},\,\it{b},\,\it{c}\,\,\it{)}}$$
Ví dụ $\it{1}$$:$
$$\it{0}\leqq \sqrt{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}}- \frac{\it{a}+ \it{b}}{\sqrt{\it{2}}}= \frac{\frac{\it{(}\,\,\it{a}- \it{b}\,\,\it{)}^{\,\it{2}}}{\it{2}}}{\sqrt{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}}+ \frac{\it{a}+ \it{b}}{\sqrt{\it{2}}}}\leqq \frac{\frac{\it{(}\,\,\it{a}- \it{b}\,\,\it{)}^{\,\it{2}}}{\it{2}}}{\frac{\it{a}+ \it{b}}{\sqrt{\it{2}}}+ \frac{\it{a}+ \it{b}}{\sqrt{\it{2}}}}= \frac{\it{(}\,\,\it{a}- \it{b}\,\,\it{)}^{\,\it{2}}}{\it{2}^{\,\frac{\it{3}}{\it{2}}}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}$$
hay$:$
$$\sqrt{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}}\leqq \frac{\it{a}+ \it{b}}{\sqrt{\it{2}}}+ \frac{\it{(}\,\,\it{a}- \it{b}\,\,\it{)}^{\,\it{2}}}{\it{2}^{\,\frac{\it{3}}{\it{2}}}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}= \frac{\it{3}\,\it{a}^{\,\it{2}}+ \it{3}\,\it{b}^{\,\it{2}}+ \it{2}\,\it{ab}}{\it{2}^{\,\frac{\it{3}}{\it{2}}}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}$$
Hoặc như bài$:$
$$\it{0}\leqq \sqrt{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}}- \left ( \it{a}+ \it{b}- \frac{\it{(}\,\,\it{4}- \it{2}^{\,\frac{\it{3}}{\it{2}}}\,\,\it{)}\it{ab}}{\it{a}+ \it{b}} \right )= \frac{\frac{\it{2}\it{(}\,\,\it{3}- \it{2}^{\,\frac{\it{3}}{\it{2}}}\,\,\it{)}\it{ab}\it{(}\,\,\it{a}- \it{b}\,\,\it{)}^{\,\it{2}}}{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,\it{2}}}}{\sqrt{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}}+ \it{a}+ \it{b}- \frac{\it{(}\,\,\it{4}- \it{2}^{\,\frac{\it{3}}{\it{2}}}\,\,\it{)}\it{ab}}{\it{a}+ \it{b}}}$$
hay$:$
$$\sqrt{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}}\leqq \frac{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\left \{ \it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\it{(}\,\,\it{3}- \it{2}^{\,\frac{\it{3}}{\it{2}}}\,\,\it{)}\it{ab} \right \}}{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\it{(}\,\,\it{2}^{\,\frac{\it{5}}{\it{2}}}- \it{5}\,\,\it{)}}$$
Chúng ta thậm chí có thể thực hiện tương tự với nhiều bài với MÀN thu nhỏ TRONG mẫu số$($làm chặt bất đẳng thức$)$$:$
$$\sqrt{\it{A}}- \it{B}= \frac{\it{A}- \it{B}^{\,\it{2}}}{\sqrt{\it{A}}+ \it{B}}= \frac{\it{A}- \it{B}^{\,\it{2}}}{\frac{\it{A}- \it{C}^{\,\it{2}}}{\sqrt{\it{A}}+ \it{C}}+ \it{B}+ \it{C}}= \frac{\it{A}- \it{B}^{\,\it{2}}}{\frac{\it{A}- \it{C}^{\,\it{2}}}{\frac{\it{A}- \it{D}^{\,\it{2}}}{\sqrt{\it{A}}+ \it{D}}+ \it{C}+ \it{D}}+ \it{B}+ \it{C}}= \,...$$
Bài viết đã được chỉnh sửa nội dung bởi DOTOANNANG: 10-04-2019 - 20:55
