Chủ đề này có 10 trả lời

[luuvanthai]

Cho các số thực dương a,b,c thỏa mãn $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=11$. Tìm GTNN của $A=(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})$

[DOTOANNANG]

$\lceil$ https://diendantoanh...ức/#entry709594 $\rfloor$

[luuvanthai]

$\lceil$ https://diendantoanh...ức/#entry709594 $\rfloor$

????    

[DOTOANNANG]

$\lceil$ Tổng quát: $\rfloor$

 

Cho các số thực $a,\,b,\,c$ dương với $\left ( a+ b+ c \right )\left ( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} \right )= 9+ m\,\,\left ( m\geqq 0,\,n\geqq 1 \right )$

 

$\text{A}_{\,n}\left ( a,\,b,\,c \right )= \left ( a^{\,n}+ b^{\,n}+ c^{\,n} \right )\left ( \frac{1}{a^{\,n}}+ \frac{1}{b^{\,n}}+ \frac{1}{c^{\,n}} \right )$

 

Với $n= 2$ thì: $\text{A}_{\,2}\left ( a,\,b,\,c \right )\geqq \text{A}_{\,2}\left ( 1,\,1,\,\frac{m+ 4+ \sqrt{m^{2}+ 8\,m}}{4} \right )= \frac{\left ( m+ 4 \right )^{2}}{2}+ 1$

 

Ta có: $\text{A}_{\,2}\left ( a,\,b,\,c \right )- 1- \frac{\left [ \text{A}_{\,1}\left ( a,\,b,\,c \right )- 5 \right ]^{\,2}}{2}= \frac{\left ( a- b \right )^{\,2}\left ( b- c \right )^{\,2}\left ( c- a \right )^{\,2}}{2\,a^{\,2}b^{\,2}c^{\,2}}\geqq 0$

 

nên ta có: $\text{A}_{\,2}= \left ( a^{\,2}+ b^{\,2}+ c^{\,2} \right )\left ( \frac{1}{a^{\,2}}+ \frac{1}{b^{\,2}}+ \frac{1}{c^{\,2}} \right )\geqq 19$

 

 

$\lceil$ https://diendantoanh...ức/#entry709594 $\rfloor$

$\lceil \,\,n= 3\,\,\rfloor$

[DOTOANNANG]

Nên ta có: $\text{A}_{\,2}= \left ( a^{\,2}+ b^{\,2}+ c^{\,2} \right )\left ( \frac{1}{a^{\,2}}+ \frac{1}{b^{\,2}}+ \frac{1}{c^{\,2}} \right )\geqq 19$

 

$\lceil \,\,m= 2\,\,\rfloor$

 

Đẳng thức xảy ra khi $\left ( a,\,b,\,c \right )= \left [ k\left ( 1,\,1,\,\left ( \frac{1+ \sqrt{5}}{2} \right )^{2} \right ) \right ]$ và các hoán vị của nó!

 

Ta có:

 

$\text{A}_{\,n}\left ( a,\,b,\,c \right )\geqq \text{A}_{\,n}\left ( 1,\,1,\,\text{M} \right )$ với $\text{M}$ là nghiệm của phương trình ${\text{M}_{\,0}}^{2}- \frac{m+ 4}{2}\,\text{M}_{\,0}+ 1= 0\,\,\Leftrightarrow \,\,\text{M}_{\,0}= \frac{m+ 4+ \sqrt{m^{\,2}+ 8\,m}}{4}$

 

hay: $\text{A}_{\,n}\left ( a,\,b,\,c \right )\geqq 5+ 2\left ( \text{M}^{\,n}+ \text{M}^{\,-\,n} \right )$

 

$\text{A}_{\,3}\left ( a,\,b,\,c \right )\geqq \frac{\left ( m+ 8 \right )\left ( m+ 4 \right )^{\,2}}{4}+ 1$

 

$\text{A}_{\,4}\left ( a,\,b,\,c\right )\geqq \frac{\left ( m^{\,2}+ m+ 8 \right )^{\,2}}{8}+ 1$

 

$\text{A}_{\,5}\left ( a,\,b,\,c \right )\geqq \frac{\left ( m+ 8 \right )\left ( m^{\,2}+ 6\,m+ 4 \right )}{16}+ 1$

[DOTOANNANG]

$\lceil$ https://diendantoanh...ức/#entry709594 $\rfloor$

$$2\left ( \text{A}_{\,3}- 1 \right )- \left ( \text{A}_{\,1}- 1 \right )\left ( \text{A}_{\,2}- 1 \right )\geqq 0$$
$\lceil$ Xin lỗi, đề sai nha! $\rfloor$

Bài viết đã được chỉnh sửa nội dung bởi DOTOANNANG: 26-11-2018 - 18:08

[DOTOANNANG]

Với $\text{A}_{\,1}\left ( a,\,b,\,c \right )= \text{N}^{\,2}\,\,\left ( \text{N}\geqq 3 \right )$ thì:

 

$$\text{A}_{\,n}\left ( a,\,b,\,c \right )\leqq \text{A}_{\,n}\left ( 1,\,\frac{-\,1+ \text{N}+ \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2},\,\frac{-\,1+ \text{N}- \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2} \right )$$

 

Với $\text{E}_{\,n}\left ( a,\,b,\,c \right )= \frac{a^{\,n}}{b^{\,n}}+ \frac{b^{\,n}}{c^{\,n}}+ \frac{c^{\,n}}{a^{\,n}},\,\text{F}_{\,n}\left ( a,\,b,\,c \right )= \frac{b^{\,n}}{a^{\,n}}+ \frac{c^{\,n}}{b^{\,n}}+ \frac{a^{\,n}}{c^{\,n}}$ thì:

 

$$\text{E}_{\,1}\left ( a,\,b,\,c \right )\leqq \text{E}_{\,1}\left ( 1,\,\frac{-\,1+ \text{N}+ \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2},\,\frac{-\,1+ \text{N}- \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2} \right )$$

 

Hoàn toàn tương tự:

 

$$\text{F}_{\,1}\left ( a,\,b,\,c \right )\geqq \text{F}_{\,1}\left ( 1,\,\frac{-\,1+ \text{N}+ \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2},\,\frac{-\,1+ \text{N}- \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2} \right )$$

 

$$\text{E}_{\,1}\left ( a,\,b,\,c \right )\geqq \text{E}_{\,1}\left ( 1,\,\frac{-\,1+ \text{N}- \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2},\,\frac{-\,1+ \text{N}+ \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2} \right )$$

[DOTOANNANG]

Với $\text{A}_{\,1}\left ( a,\,b,\,c \right )= \text{N}^{\,2}\,\,\left ( \text{N}\geqq 3 \right )$ thì:

 

$$\text{A}_{\,n}\left ( a,\,b,\,c \right )\leqq \text{A}_{\,n}\left ( 1,\,\frac{-\,1+ \text{N}+ \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2},\,\frac{-\,1+ \text{N}- \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2} \right )$$

 

Với $\text{E}_{\,n}\left ( a,\,b,\,c \right )= \frac{a^{\,n}}{b^{\,n}}+ \frac{b^{\,n}}{c^{\,n}}+ \frac{c^{\,n}}{a^{\,n}},\,\text{F}_{\,n}\left ( a,\,b,\,c \right )= \frac{b^{\,n}}{a^{\,n}}+ \frac{c^{\,n}}{b^{\,n}}+ \frac{a^{\,n}}{c^{\,n}}$ thì:

 

$$\text{E}_{\,1}\left ( a,\,b,\,c \right )\leqq \text{E}_{\,1}\left ( 1,\,\frac{-\,1+ \text{N}+ \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2},\,\frac{-\,1+ \text{N}- \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2} \right )$$

 

Hoàn toàn tương tự:

 

$$\text{F}_{\,1}\left ( a,\,b,\,c \right )\geqq \text{F}_{\,1}\left ( 1,\,\frac{-\,1+ \text{N}+ \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2},\,\frac{-\,1+ \text{N}- \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2} \right )$$

 

$$\text{E}_{\,1}\left ( a,\,b,\,c \right )\geqq \text{E}_{\,1}\left ( 1,\,\frac{-\,1+ \text{N}- \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2},\,\frac{-\,1+ \text{N}+ \sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2} \right )$$

 

Từ trên, ta được: 

 

$$\frac{-\,3+ \text{N}^{2}- \left ( \text{N}- 3 \right )\sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2}\leqq \left [ \text{E}_{\,1},\,\text{F}_{\,1} \right ]\left ( a,\,b,\,c \right )\leqq \frac{-\,3+ \text{N}^{\,2}+ \left ( \text{N}- 3 \right )\sqrt{\left ( \text{N}- 3 \right )\left ( \text{N}+ 1 \right )}}{2}$$

 

$$\Leftrightarrow \,\,\left [ \text{E}_{\,1}\left ( a,\,b,\,c \right )- \text{F}_{\,1}\left ( a,\,b,\,c \right ) \right ]^{^{2}}+ \left [ \text{E}_{\,1}\left ( a,\,b,\,c \right )+ \text{F}_{\,1}\left ( a,\,b,\,c \right ) \right ]^{^{2}}\leqq \left ( -\,3+ \text{N}^{\,2} \right )^{2}+ \left ( \text{N}- 3 \right )^{3}\left ( \text{N}+ 1 \right )$$

 

$$\Leftrightarrow {\text{E}_{1}}^{2}\left ( a,\,b,\,c \right )+ {\text{F}_{1}}^{2}\left ( a,\,b,\,c \right )\leqq \frac{1}{2}\left [ \left ( -\,3+ \text{N}^{\,2} \right )^{2}+ \left ( \text{N}- 3 \right )^{3}\left ( \text{N}+ 1 \right ) \right ]$$

 

Ở bài trên gốc $\lceil\,\,\text{N}^{2}= 11\,\,\rfloor$ thì:

 

$\left ( \frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \right )^{2}+ \left ( \frac{b}{a}+ \frac{c}{b}+ \frac{a}{c} \right )^{2}\leqq 178- 44\,\sqrt{11}\,\,$ $\Rightarrow \,\,\sum\limits_{cyc}\,\frac{a^{\,2}}{b^{\,2}}+ \sum\limits_{cyc}\,\frac{b^{\,2}}{a^{\,2}}\leqq 16\,\,\Rightarrow \,\,\left ( \sum\limits_{cyc}\,a^{\,2} \right )\left ( \sum\limits_{cyc}\,\frac{1}{a^{\,2}} \right )\leqq 19$

[DOTOANNANG]

Với $\text{A}_{\,1}\left ( a,\,b,\,c \right )= \text{N}^{\,2}\,\,\left ( \text{N}\geqq 3 \right )$ thì:

 

$\text{A}_{\,2}\left ( a,\,b,\,c \right )\leqq \left [ \text{N}\left ( \text{N}- 2 \right ) \right ]^{2}$

 

$\text{A}_{\,3}\left ( a,\,b,\,c \right )\leqq \left ( \text{N}^{\,3}- 3\,\text{N}^{\,2}+ 3 \right )^{^{2}}$

 

$\text{A}_{\,4}\left ( a,\,b,\,c \right )\leqq \left [ \text{N}\left ( \text{N}- 2 \right )\left ( \text{N}^{\,2}- \text{N}- 2 \right ) \right ]^{\,2}$

Bài viết đã được chỉnh sửa nội dung bởi DOTOANNANG: 25-11-2018 - 14:38

[DOTOANNANG]

Cách khác với $m= 1,\,n= 2$ : $\lceil$ https://diendantoanh...72/#entry702543 $\rfloor$ 

 

 

$\lceil$ Tổng quát: $\rfloor$

 

Cho các số thực $a,\,b,\,c$ dương với $\left ( a+ b+ c \right )\left ( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} \right )= 9+ m\,\,\left ( m\geqq 0,\,n\geqq 1 \right )$

 

$\text{A}_{\,n}\left ( a,\,b,\,c \right )= \left ( a^{\,n}+ b^{\,n}+ c^{\,n} \right )\left ( \frac{1}{a^{\,n}}+ \frac{1}{b^{\,n}}+ \frac{1}{c^{\,n}} \right )$

 

Với $n= 2$ thì: $\text{A}_{\,2}\left ( a,\,b,\,c \right )\geqq \text{A}_{\,2}\left ( 1,\,1,\,\frac{m+ 4+ \sqrt{m^{2}+ 8\,m}}{4} \right )= \frac{\left ( m+ 4 \right )^{2}}{2}+ 1$

 

Ta có: $\text{A}_{\,2}\left ( a,\,b,\,c \right )- 1- \frac{\left [ \text{A}_{\,1}\left ( a,\,b,\,c \right )- 5 \right ]^{\,2}}{2}= \frac{\left ( a- b \right )^{\,2}\left ( b- c \right )^{\,2}\left ( c- a \right )^{\,2}}{2\,a^{\,2}b^{\,2}c^{\,2}}\geqq 0$

 

nên ta có: $\text{A}_{\,2}= \left ( a^{\,2}+ b^{\,2}+ c^{\,2} \right )\left ( \frac{1}{a^{\,2}}+ \frac{1}{b^{\,2}}+ \frac{1}{c^{\,2}} \right )\geqq 19$

Bài viết đã được chỉnh sửa nội dung bởi DOTOANNANG: 25-11-2018 - 18:27

[DOTOANNANG]

Giả sử rằng:

 

$\text{V}_{\,k}\left ( x_{\,1},\,x_{\,2},\,...,\,x_{\,n} \right )= \left ( {x_{\,1}}^{k}+ {x_{\,2}}^{k}+ \,...\,+ {x_{\,n}}^{k} \right )\left ( \frac{1}{{x_{\,1}}^{k}}+ \frac{1}{{x_{\,2}}^{k}}+ \,...\,+ \frac{1}{{x_{\,n}}^{k}} \right )$ , $x_{\,i}> 0\,\,\,\,\,\,\,\,\lceil \,\,i= \overline{1,\,n} \,\,\rfloor$

 

Khi đó:

 

$$\sqrt{\text{V}_{\,2}\left ( x_{\,1},\,x_{\,2},\,...,\,x_{\,n} \right )}\leqq \sqrt{\text{V}_{\,1}\left ( x_{\,1},\,x_{\,2},\,...,\,x_{\,n} \right )}\left [ \sqrt{\text{V}_{\,1}\left ( x_{\,1},\,x_{\,2},\,...,\,x_{\,n} \right )}- n+ 1 \right ]$$